Can You Conclude That This Parallelogram Is a Rhombus? Explain.

ou've figured out the solution to the problem — fantastic! But you're not finished. Whether you lot are writing solutions for a competition, a periodical, a message lath, or just to show off for your friends, you must chief the art of communicating your solution conspicuously.

Brilliant ideas and innovative solutions to bug are pretty worthless if you can't communicate them. In this article, we explore many aspects of how to write a clear solution. Beneath is an index; each page of the article includes a sample 'How Not To' solution and 'How To' solution. One common theme you'll find throughout each point is that every fourth dimension you make an experienced reader have to call back to follow your solution, you lose.

As y'all read the 'How To' solutions, you may think some of them are overwritten. Indeed, some of them could be condensed. Some steps we chose to prove could probably exist cited without proof. Still, it is far improve to prove also much too clearly than to evidence too little. Rarely will a reader complain that a solution is as well easy to understand or besides easy on the centre.

1 note of warning: many of the problems we utilise for examples are extremely challenging problems. Beginners, and even intermediate students, should not be upset if they take difficulty solving the bug on their ain.

Tabular array of Contents:

• Take a Plan
• Readers Are Not Interpreters
• U s due east South p a c e
• sdrawkcaB knihT, Write Forwards
• Proper noun Your Characters
• A Movie is Worth a Thousand Words
• Solution Readers, Not Listen Readers
• Follow the Lemmas
• Articulate Casework
• Proofreed
• Bookends

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Have A Programme

Your goal in writing a clear solution is to prevent the reader from having to think. You must limited your ideas clearly and concisely. The experienced reader should never take to wonder where you are headed, or why any merits you brand is truthful. The first step in writing a clear solution is having a plan. Make a simple outline of your solution. Include the items y'all'll demand to define, and the club in which you will write upward the important parts of your solution. The outline will help ensure that you don't skip annihilation and that you put your steps in an order that's easy to follow.

Here'due south a sample problem:

A sphere of radius 𝑟 is inscribed in a tetrahedron. Planes tangent to this sphere and parallel to the faces of the tetrahedron cut off four small tetrahedra from the tetrahedron; these small tetrahedra have inscribed spheres with radii 𝑎, 𝑏, 𝑐, 𝑑. Show that:

𝑎+𝑏+𝑐+𝑑=ii𝑟

Here's a solution that looks short but is pretty tough to read.

How Not to Write the Solution:

Allow our tetrahedron be \(ABCD\). The small tetrahedron which includes vertex A is similar to the big tetrahedron. Since the confront of this tetrahedron parallel to face \(BCD\) is tangent to the sphere inscribed in \(ABCD\), the distance between \(BCD\) and this parallel face up of the small tetrahedron is \(2r\). Let's call that small tetrahedron \(AXYZ\). Hence, the distance from \(A\) in \(AXYZ\) is \(h_a - 2r\), where \(h_a\) is the length of the distance from \(A\) to side \(BCD\). Therefore the ratio of the altitudes from \(A\) in \(AXYZ\) and \(ABCD\) is \((h_a- 2r)/h_a\). Since these two tetrahedrons are like with ratio \(a/r\) (since that'southward the ratio of the corresponding lengths, namely the radii of the inscribed spheres) nosotros have \(a/r =\) \((h_a- 2r)/h_a\). The volume of the tetrahedron is \([A]h_a/iii\), where \([A]\) is the area of triangle \(BCD\). The volume of the tetrahedron can too be written \(rS/three\), where \(Southward\) is the surface area of \(ABCD\). We can testify that by letting \(I\) be the center of the inscribed sphere. Then the volume of the tetrahedron is the sum of the volumes of the tetrahedra \(IABC\), \(IABD\), \(IBCD\), and \(IACD\). The volume of \(IABC\) is \(r[D]/3\), where \([D]\) is defined like we defined \([A]\) above. We can similarly detect the volumes of the other 4 pieces. When we add them all up, nosotros get \[ \text{Book of } ABCD = ([A] + [B] + [C] + [D]) r/iii = rS/3.\] Nosotros set up that equal to our other volume expression and become \(h_a= rS/[A]\). If we rearrange our equation from above, we have \(a = r - 2r^2/ h_a\). We can then put in the \(h_a\) expression we just found to go: $$a = r - 2r[A]/S.$$  If we ascertain \([B]\), \([C]\), and \([D]\) merely like nosotros divers \([A]\), we tin can use the same argument to go: $$b = r - 2r[B]/S,$$$$c = r - 2r[C]/Southward,$$$$d = r - 2r[D]/South.$$ Adding these and our expression for \(A\), we go: $$a + b + c + d = 4r - 2r\cdot([A]+[B]+[C]+[D])/S = 2r,$$ equally desired.

(Full general solution method found by community member zabelman in the Olympiad Geometry form.)

The main problem with the above solution is 1 of organization. We defined variables later they popped upwards. Midway through the solution we sidetracked to show the volume of ABCD is 𝑟𝑆/3. Sometimes we wrote of import equations right in our paragraphs instead of highlighting them by giving them their own lines.

If nosotros outline before writing the solution, we won't have these problems. We tin list what we need to ascertain, decide what items nosotros need to prove before our principal proof (nosotros call these lemmas), and list the important steps so we know what to highlight.

Our scratch sheet with the outline might have the following:

Stuff to ascertain: 𝐴𝐵𝐶𝐷,ℎ𝑎,𝑆,[𝐴],𝐴𝑋𝑌𝑍.

Order of things to evidence:

1. Volume \(ABCD = rS/3\) (lemma)
ii. Evidence altitude \(AXYZ = h_a - 2r\)
three. Use similarity to get \(a = r - 2r^2/h_a\)
four. Equate volumes to go \(1/h_a= A/(rS)\)
5. sub \(4\) into \(3\) and add

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This list looks obvious once y'all have it written up, just if you just plough ahead with the solution without planning, yous may end up skipping items and having to wedge them in as we did in our 'How Non to Write the Solution'.

How to Write the Solution:

Permit our original tetrahedron be \(ABCD\). We define: \([A]\) = the surface area of the face of \(ABCD\) opposite \(A\) \(h_a\) = the length of the distance from \(A\) to \(BCD\) \(S\) = the surface area of \(ABCD\) \(AXYZ\) = i of the small tetrahedrons formed as described

Ascertain \([B]\), \([C]\), \([D]\) and \(h_b\), \(h_c\), \(h_d\) similarly.

Lemma: The volume of tetrahedron \(ABCD\) is given by \(rS/3\).
Proof: Let \(I\) be the middle of the inscribed sphere. The volume of \(ABCD\) is the sum of the volumes of the tetrahedra \(IABC\), \(IABD\), \(IBCD\), and \(IACD\). The volume of \(IABC\) is \((r)[D]/3\), since the altitude from \(I\) to \(ABC\) is a radius of the inscribed sphere of \(ABCD\). We tin can similarly find the volumes of the other iv pieces. Adding these 4 tetrahedra gives us:
\[ \text{Volume of } ABCD = ABCD = ([A] + [B] + [C] + [D])r/three = rS/3\]

as desired.

Since face up \(XYZ\) of small tetrahedron \(AXYZ\) is parallel to face \(BCD\), tetrahedron \(AXYZ\) is like to \(ABCD\). The ratio of corresponding lengths in these tetrahedra equals the ratio of the radii of their inscribed spheres, or \(a/r\). Since \(XYZ\) is tangent to the sphere inscribed in \(ABCD\), the altitude between \(BCD\) and \(XYZ\) is \(2r\). Hence, the altitude from \(A\) to \(XYZ\) is \(h_a - 2r\). Therefore the ratio of the altitudes from \(A\) in the two tetrahedra is \((h_a - 2r)/h_a\). Hence, \(a/r = (h_a - 2r)/h_a\), or \(a = r - 2r^two/h_a\). The volume of the tetrahedron is \(h_a[A]/iii\). Setting this equal to the expression from Lemma 1 yields: \(h_a = rS/[A]\), and substituting this into equation (one), we get: \(a = r - 2r[A]/S\). By the aforementioned argument, nosotros take: $$b = r - 2r[B]/Due south,$$$$c = r - 2r[C]/S,$$$$d = r - 2r[D]/Due south.$$ Adding these and our expression for \(A\), we become: $$a + b + c + d = 4r - 2r\cdot ([A]+[B]+[C]+[D])/S = 2r,$$ every bit desired.

(General solution method found past community member zabelman in the Olympiad Geometry form.)

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Readers Are Not Interpreters

The first thing a reader sees on your paper isn't the construction of your solution. It isn't the answer, it isn't the words you choose. It'due south how the solution sits on the paper. If the reader has to decipher scrawl, y'all're going to lose him. Ideally, you lot'll typeset your solution with a program like LaTeX. However, in most contests you lot don't have the luxury of turning to a computer and you'll accept to write information technology out by hand. There are few very important rules of thumb when writing a solution past hand. Many are obvious, some are less so. You lot should follow them all.

1. Use blank paper. Don't utilize graph newspaper or lined paper – the lines oft make solutions harder to read. Never apply newspaper that is torn out of a spiral notebook.
2. Respect margins. If you lot are starting with a completely bare piece of newspaper, draw the margins on all four sides (top, lesser, right, left). Make your margins at least 0.5 inches, and preferably a total inch.
3. Write horizontally; never plough your writing when you accomplish the end of a line in club to jam in a fiddling more information. You tin always start a new line or a new page.
4. Leave space at the tiptop for a 'Page _ of _' so the reader knows how many pages at that place are, and what page she'southward on. You probably won't know how many pages you'll write when you get-go, but you can fill these out when yous're finished. If you become to the lesser of a page and your solution must keep on another folio, write 'Continued' at the bottom of that page and then the reader knows we're non finished. (This also helps readers know if they're missing pages.)
5. Don't write in cursive. Print. And print clearly.
6. Use pen. If you must use pencil, practice not erase – the smudges from erasers make a mess.
7. When you make a mistake y'all'd like to omit, draw a single line through it and move on. If it'south a large cake to omit, draw an 'X' through it and move on. Don't scribble out big blocks of text.
8. If yous left something out and want to add together it at the end, put a elementary symbol, like a (*), at the point where you would similar the new text to exist considered added, and leave a brief note, such every bit 'Proof below.' Beneath, y'all tin write '(*) Addendum:' and proceed with the proof. Don't use a bunch of arrows to direct the reader all over the page.

Problem: Let \(Due south(north)\) be the sum of the digits of \(north\). Find $$S(S(S(4444^{4444}))).$$

Below are two solutions. Neither solution is picture-perfect; when you lot're under fourth dimension pressure, it'south difficult to write perfect-looking proofs. You should notice the 2nd ane much more enjoyable to read. When y'all're writing solutions, keep the higher up tips in listen, and simply remember, 'If they can't read information technology, information technology's non correct.'

How Non to Write the Solution:

That solution above is a mess. The one below took me just as long to write, and is much easier to read.

How to Write the Solution:

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U south e S p a c e

imagineyoutrytoreadaparagraphoftextthathasnopuncuationnocapitalsandjustenough
spaceinittobreakuplinessoitdoesntmessupbrowsersitsreallytoughtoreadandpretty
soonyoulldecidethatitsnotworthreadingandyoullgoandreadsomethingelseyouwont
realizehowterriblyharditistotypelikethisitshardbecausewhenyoureusedtowritingcle
arlyandusingspaceandpunctuationandsentencestructureitgetsreallyhardtowritewith
outitsimilarlyonceyougetusedtoproperlyusingspaceinwritingyoursolutionsitwillbesec
ondnatureandyoullactuallyfindithardtowriteanindecipherableproof

When you lot write your solution you should:

1. Give each important definition or equation its ain line.
2. Don't bury too much algebra in a paragraph. You tin write line later on line of algebra, just put each step on its own line. Don't cram the algebra in a paragraph.
3. Characterization equations or formulas or lemmas or cases you lot will apply after very clearly.
4. Remember that at that place's ever more paper.

Problem: Let \(p(x)\) exist a polynomial with degree \(98\) such that \(p(n) = 1/northward\) for \(n =\) \(ane, 2, three,\) \(4, \ldots , 99\). Determine \(p(100)\).

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Have fun reading this solution.

How Not to Write the Solution:

Let \(r(x) =\) \(x (p(x) - 1/x)\) \(= x p(x) - i.\) Since \(p(x)\) is a polynomial with degree \(98\), \(r(10)\) is a polynomial with degree \(99\). Since \(r(ten) =\) \(ten (p(x) - 1/x),\) and we are given that \((p(ten) - 1/x) = 0\) for \(10 = 1,\) \(2, 3, \ldots , 99\), \(r(ten)\) has roots \(one, 2, \ldots, 99.\) Since \(r(x)\) has degree \(99\), these are the simply roots of \(r(x)\), which must thus have the form \(r(x) =\) \(c(10 - 1)(x - ii)\) \((x - 3) \cdots (10 - 99)\) for some abiding \(c\). To find \(c,\) we first let \(10 = 0\) in equation \(r(x) =\) \(x p(x) - 1,\) yielding \(r(0) = -one.\) Letting \(x = 0\) in \(r(x) =\) \(c(ten - 1)(x - two)\) \((x - iii) \cdots (x - 99)\) yields \(r(0) = -c(99!);\) hence, \(c = one/99!.\) Thus, we have \(r(x) =\) \((x - ane)(ten - 2)\) \((x - iii) \cdots (x - 99)/99!.\) We tin combine the equations \(r(ten) =\) \(10 p(x) - 1\) and \(r(10) =\) \((x - 1)(10 - 2)\) \((x - 3) \cdots (ten - 99)/99!\) and let \(ten = 100\) to find \(100p(100) - i =\) \( (100 - ane)(100 - 2)\) \((100 - three) \ldots(100 - 99)/99!,\) and then \(100p(100) - 1 =\) \( 99!/99! = i,\) and so \(p(100) = 1/50.\)

Hither's the same solution, with almost the same wording.

How to Write the Solution:

Let \[ r(ten)=x(p(ten)-one/10)=xp(x)-1.\tag{i} \] Since \(p(x)\) is a polynomial with degree \(98\), \(r(x)\) is a polynomial with degree \(99\). Since \(r(10) =\) \(ten (p(x) - ane/10)\), and we are given that \((p(x) - 1/10) = 0\) for \(x = \) \(ane, 2,\) \(3, \ldots, 99,\) \(r(x)\) has roots \(1, ii, \ldots , 99.\) Since \(r(x)\) has degree \(99\), these are the but roots of \(r(x)\), which must thus accept:  \[ r(x)=c(x-1)(x-two)(x-3)\cdots(x-99) \tag{ii} \] for some constant \(c\). To find \(c\), nosotros first let \(x = 0\) in equation \((ane)\), yielding \(r(0) = -1.\) Letting \(ten = 0\) in \((2)\) yields \(r(0) = -c(99!)\); hence, \(c = 1/99!\). Thus, we have:  $$\begin{equation}r(x)=(x-one)(10-2)(10-3)\cdots(ten-99)/99 ! \tag{3}\terminate{equation}$$ We can combine equations \((1)\) and \((iii)\) and let \(ten = 100\) to find \[100p(100) - 1 = (100 - 1)(100 - 2)(100 - 3) \cdots (100 - 99)/99!\]\[100p(100) - 1 = 99!/99! = 1\]\[p(100) = 1/50.\]

Which would you rather read?

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sdrawkcaB knihT, Write Forward

The following is an excerpt from a cookbook that was never written:

"Figuring out how to make an omelette is piece of cake. Anybody who has eaten an omelette knows that an omelette is typically fabricated with several eggs filled with various foods such as ham, peppers, onions, and bacon and is often cooked with cheese. The fact that all these ingredients cease upwardly inside the egg means that nosotros should begin cooking the eggs flatly on a pan and and so add the ingredients. We can then roll role of the egg over the ingredients so equally to trap them on the inside. If we needed some of the ingredients precooked nosotros could do that before adding them to the eggs…"

It is one thing to figure out how to make an omelette. Information technology is another to explain to somebody else how to brand i. Starting our caption from the beginning is much clearer than starting with the finished omelette.

"Prepare vegetables and other desired omelette fillings. Beat eggs. Start cooking the eggs. Add your fillings in the middle then that part of the egg can be pulled over the ingredients. When the omelette is closed, continue to melt and flip the omelette until the eggs await well-cooked."

The reader doesn't care how the process of cooking an omelette was unraveled past the author. The reader just wants to know how to make an omelette.

Think of solutions as recipes. Beginning at the commencement and move forward. List the ingredients and explain how and when to add them to the pot.

Hither'south a sample trouble.

Problem: Allow \(a\), \(b\), and \(c\) denote the lengths of the sides of a triangle. Show that $$a^2(-a + b + c) + b^2(a - b + c) + c^2(a + b - c)\leq 3abc.$$

This solution might be a skillful way to see how we might come up up with a solution from scratch, but information technology's not a particularly well-written proof:

How Not to Write the Solution:

We note that the inequality contains the factors \((-a + b + c)\), \((a - b + c)\), and \((a + b - c)\). These factors point to using the triangle inequality so it seems natural to leave the factors alone and invoke the fact that each is nonnegative. Since each of these three factors is multiplied by the square of the length of a side information technology might exist possible to dispense the inequality into something involving these nonnegative triangle inequality factors multiplied by perfect squares. We could then argue that this sum must also be nonnegative. Nosotros begin by moving \(3abc\) to the left hand side: $$a^2(-a + b + c) = b^ii(a - b + c) + c^ii (a + b - c) - 3abc \le 0.$$ If nosotros were to view \(3abc\) as the sum of \(3\) terms that are each the product of \(ab\), \(bc\), or \(ca\) and one of the triangle inequality factors, we begin to get an idea every bit to how the inequality tin exist reorganized. Since the inequality is cyclic, it seems natural to take these products in a mode that preserves the cyclic nature. For instance, nosotros multiply \(ab\) with \((a + b - c)\) because \(a\) and \(b\) have the same sign in \((a + b - c)\):  \begin{align*} ab( a + b - c) &= a^2b +ab^2 -abc \ge 0, \\ ab (-a + b +c) &= -abc + b^2c + bc^2 \ge 0, \\ ca (a - b + c) &=a^2c -abc + air-conditioning^ii \ge 0. \end{align*} We come across the \(-3abc\) in the sum of these products. Examining the other terms in \[ab(a + b - c) = a^2b + ab^2 - abc,\] we notice that \(a^2b + ab^2\) are factors that would pop out of \((a - b)^2(a + b - c) \). Expanding the squared parts of expressions like \((a-b)^2(a+b-c)\) we get  \begin{marshal*} (a - b)^two(a + b - c) &= a^2(a + b - c) - 2ab(a + b - c) + b^2(a + b - c) \ge 0, \\ (b - c)^2(-a + b + c) &= b^2(-a + b + c) - 2bc(-a + b + c) + c^2(-a + b + c) \ge 0, \\ (c - a)^ii(a - b + c) &= c^2(a - b + c) - 2ca(a - b + c) + a^2(a - b + c) \ge 0.\finish{marshal*} Adding these inequalities together we begin to see the inequality take shape: \begin{align*} &a^2(a + b - c + a - b + c) + b^2(a + b - c - a + b + c) + c^two(-a + b + c + a - b + c) \\ &- 2a^2b - 2ab^2 + 2abc + 2abc - 2b^2c - 2bc^2 - 2a^2c + 2abc - 2ac^ii  \\ &= a^2(2a - 2b - 2c) + b^2(-2a + 2b - 2c) + c^2(-2a - 2b + 2c) + 6abc \ge 0. \end{marshal*} Multiplying this inequality by \(-one/2\) reverses the inequality sign and gives us $$a^ii(-a + b + c) + b^two(a - b + c) + c^ii(a + b - c) - 3abc \ge 0.$$ Adding \(3abc\) to both sides gives u.s.a. $$a^two(-a + b + c) + b^two(a - b + c) + c^ii(a + b - c) \ge 3abc$$ and we are done.

The cookbook manner is easier to read and far more convincing.

How to Write the Solution:

Co-ordinate to the triangle inequality, the sum of any two sides of a triangle is at least as smashing as the length of the third side. Thus, we accept the inequalities \begin{marshal*} a - b - c&\leq0, \\ b - c - a&\leq0, \\ c - a - b&\leq0. \end{align*} Multiplying these by perfect squares leave each left-hand side nonpositive, so \begin{align*} (b - c)^2(a - b - c)&\leq0, \\ (c - a)^2(b - c - a)&\leq0, \\ (a - b)^2(c - a - b)&\leq0, \terminate{align*} or \begin{align*} (b^2 - 2bc + c^2)(a - b - c)&\leq0, \\ (c^2- 2ac + a^2)(b - c - a) &\leq 0, \\ (a^two - 2ab + b^2)(c - a - b)&\leq0. \end{marshal*} Adding these inequalities we get \brainstorm{align*} &a^2(a + b - c + a - b + c) + b^2(a + b - c - a + b + c) + c^ii(-a + b + c + a - b + c) \\ &- 2a^2b - 2ab^2 + 2abc + 2abc - 2b^2c - 2bc^2 - 2a^2c + 2abc - 2ac^ii  \\ &= a^2(2a - 2b - 2c) + b^2(-2a + 2b - 2c) + c^ii(-2a - 2b + 2c) + 6abc \ge 0. \end{align*} Adding \(6abc\) to both sides and dividing by \(two\) nosotros take the desired $$a^2(-a + b + c) + b^two(a - b + c) + c^2(a + b - c)\leq3abc.$$

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Name Your Characters

A large thin-shelled vehicle for a young fowl that was created by a huge female bird sat on a wall. The large thin-shelled vehicle for a young fowl that was created by a huge female bird had a great fall. All the horses of the great man who lived in a large castle that ruled over the people in the country and all the men of the nifty human who lived in a large castle that ruled over the people in the state couldn't put the large thin-shelled vehicle for a immature fowl that was created past a huge female person bird back together again.

Proofs are a lot like stories. When writing a solution your job is tell a math story in a way your audition volition understand and savor. Instead of writing most 'A big thin-shelled vehicle for a young fowl that was created by a huge female bird,' we telephone call that big egg 'Humpty-Dumpty' and tell the story. Likewise, a well-written proof ofttimes involves naming the important quantities or ideas that play a role in the story of your solution. Naming your characters can as well help you find solutions to issues, and then it's non something you should wait until proof-writing time to do.

When you do name your characters, you name them just, clearly, and write up front, so the reader knows exactly where to become to observe out exactly who this 𝑛 person is and what that 𝑓(𝑥) function stands for.

Here'south an example trouble.

Problem: Testify that for whatever prepare of \(100\) integers that there is some subset such that the sum of its elements is a multiple of \(100\).

The solution below is hard to read considering the integers and the sums that are the key to the solution remain unnamed.

How Not to Write the Solution:

Suppose we put the numbers in our set in some fixed gild. If we start from the beginning, there are \(100\) sums we can make past but adding up starting from the beginning number. We could add up the first \(2\) numbers, or the first \(4\), the first \(57\), or whatever. If one of these sums is a multiple of \(100\), then we are done. If none of the sums is a multiple of \(100\), then we need to consider the remainders when each of these terms is divided by \(100\). There are \(100\) total remainders since there are \(100\) sums. Since none of them is \(0\), in that location are at almost \(99\) different remainders among these \(100\) remainders. Therefore, two of these remainders must exist the same since if in that location weren't at least two the same in that location could only be \(99\) total remainders (since nosotros know none is zero). Now, take the difference between the 2 sums which have the aforementioned remainder when divided by \(100\). This deviation must have a residue of \(0\) when divided by \(100\). Suppose we have subtracted the sum with fewer numbers from the one with more numbers. When nosotros take this divergence, all the numbers in the second sum cancel with numbers in the first sum, because each sum is only calculation upwards numbers in our set starting with the beginning one but the second sum is shorter. Due to this cancellation, the deviation of these two sums which have the aforementioned remainder when divided by \(100\) results in a sum of numbers in the original set. We have shown that this difference has a remainder of \(0\) when divided past \(100\), so this is our desired sum of numbers in the set that are divisible by \(100\).

The solution below is easy to read because the main characters have names. Specifically, nosotros name the integers in the set and the sums of the elements in subsets that we examine. These names allow us to follow the characters throughout the story. They also let the writer to describe the characters more completely and succinctly.

How to Write the Solution:

Telephone call the 100 integers \(n_1,\) \(n_2,\) \(\ldots,\) \(n_{100}\). Let \(S_k = n_1 + n_2 + \dots + n_k\) for \(thousand = 1,\) \(ii, \dots, 100.\) Case 1: If \(S_1, S_2,\dots, S_{100}\) are all distinct (modernistic 100), then exactly one of them must exist a multiple of 100. Case 2: Otherwise, the 100 sums, \(S_k\), have at most 99 distinct residues (modern 100) and by the Pigeonhole Principle two of the sums, \(S_k\), take the aforementioned residue (mod 100). Thus means there be some integers \(j\) and \(k\), \(0 < j < k < 101\), such that $$S_k \equiv S_j \pmod{100};$$ thus, $$S_k \equiv 0 \pmod{100}.$$ Now, consider the subset with elements \(n_{j+1},\) \(n_{j+two},\) \(\ldots,\) \(n_k\). The sum of the elements of this subset is \brainstorm{marshal*} & n_{j+i} + n_{j+2} + \dots + n_k \\ & = (n_1 + n_2 + \dots + n_k) - (n_1 + n_2 + \dots + n_j) \\ S_k& \equiv 0 \pmod{100}. \end{align*} Thus this sum is a multiple of 100 and we are done.

A Picture is Worth a Grand Words

When you're writing a solution to a geometry problem, or any problem involving a picture, you lot should include the diagram. If you don't include the diagram, you often make the grader have to depict it for you. Even if the diagram is given in the problem, y'all should include information technology in your solution. If you make your reader go looking somewhere else for a diagram, you are very likely to lose their attention.

Draw your diagram precisely. Employ a geometry rendering program if yous are typesetting your solution, or use a ruler and compass if you lot are writing your solution past hand.

Here's an example.

Trouble: A point \(D\) is placed on side \(BC\) of triangle \(ABC\). Circles are inscribed in \(ABD\) and \(ACD\). Their mutual exterior tangent (other than \(BC)\) meets \(Advertisement\) at \(K\). Prove that the length of \(AK\) is contained of \(D\).

Here's a solution without a diagram.

How to Write the Solution:

Let our circles be \(O\) and \(O'\). Allow \(Grand\) and \(M'\) exist on \(O\) and \(O'\) such that \(MM'\) is the common tangent through \(K\). Let \(50\) and \(L'\) exist the points where \(Ad\) meets circles \(O\) and \(O'\), respectively. Permit \(N\) and \(N'\) be the points where \(BC\) meets circles \(O\) and \(O'\), respectively. We will show that \(AK = \) \({(AB + AC - BC)}/2\), and thus prove that the length of \(AK\) is independent of \(D\). Since tangents from a bespeak to a circle are equal, we have both \(DN = DL\) and \(DN' = DL'\). Thus, $$NN' = DN + DN' = DL + DL' = 2DL + LL'.$$ Similarly, nosotros have $$MM' = MK + KM' = KL + KL' = 2KL' + LL'.$$ Since \(MM' = NN'\) by symmetry, we conclude that \(KL = DL\). Hence, $$MM' = NN' = DL + LL' + KL' = KD.$$ We can compute \(NN'\), and hence \(KD\), in terms of \(Advertisement\) and the sides of the triangle: \begin{align*} DN &= {(AB + Ad + BD)}/2 - AB \\ DN' &= {(AC + AD + CD)}/ii - AC \cease{marshal*} Thus, $$NN' = ND + DN' = Ad + {BC}/2 - {Ac}/ii - {AB}/2.$$ Since \(NN' = KD\), we take $$AD - KD = AK = {(Ac + AB - BC)}/2,$$ as desired. Since \(A\), \(B\), and \(C\) are independent of \(D\), we conclude that the length \(AK\) is independent of \(D\).

(Solution method found by community fellow member 3cnfsat in the Olympiad Geometry course)

Here'south a solution that includes the diagram:

How to Write the Solution:

Allow our circles be \(O\) and \(O'\). Permit \(M\) and \(M'\) be on \(O\) and \(O'\) such that \(MM'\) is the common tangent through \(K\). Let \(50\) and \(L'\) exist the points where \(AD\) meets circles \(O\) and \(O'\), respectively. Allow \(N\) and \(N'\) be the points where \(BC\) meets circles \(O\) and \(O'\), respectively. We will show that \(AK = \) \({(AB + AC - BC)}/2\), and thus show that the length of \(AK\) is contained of \(D\). Since tangents from a point to a circle are equal, nosotros have both \(DN = DL\) and \(DN' = DL'\). Thus, $$NN' = DN + DN' = DL + DL' = 2DL + LL'.$$ Similarly, we have $$MM' = MK + KM' = KL + KL' = 2KL' + LL'.$$ Since \(MM' = NN'\) past symmetry, we conclude that \(KL = DL\). Hence, $$MM' = NN' = DL + LL' + KL' = KD.$$ We tin can compute \(NN'\), and hence \(KD\), in terms of \(AD\) and the sides of the triangle: \begin{align*} DN &= {(AB + AD + BD)}/2 - AB \\ DN' &= {(AC + Advertizing + CD)}/ii - AC \end{align*} Thus, $$NN' = ND + DN' = Advertizement + {BC}/2 - {Air conditioning}/ii - {AB}/ii.$$ Since \(NN' = KD\), we have $$Advert - KD = AK = {(AC + AB - BC)}/2,$$ equally desired. Since \(A\), \(B\), and \(C\) are independent of \(D\), we conclude that the length \(AK\) is contained of \(D\).

(Solution method found past community member 3cnfsat in the Olympiad Geometry grade)

In our solution above, we used the fact that the length of the segments from a vertex of a triangle (similar vertex \(D\) of triangle \(ABD\) above) to the points of tangency of the inscribed circle with the sides of the triangle from that vertex (segments \(DN\) and \(DL\) higher up) equals half the perimeter of the triangle minus the opposite side of the triangle. Applying this principle to find length \(DN\) in triangle \(ABD\) gives us: $$DN = (AB + AD + BD)/two - AB$$

If you aren't familiar with this fact, try to prove it yourself (and write a nice solution). Every adept geometer reaches for this fact equally easily equally they achieve for the Pythagorean Theorem.

Solution Readers, non Mindreaders

A full solution does not but hateful a correct respond. You should justify every notable step of your solution. An experienced reader should never wonder 'Why is that true?' while reading your solution. She should also never be left in doubt equally to whether or non you know why it is true.

It'due south not always clear what steps yous tin can presume the reader understands and what steps yous accept to explain. Hither area few guidelines:

1. If you can cite a theorem that has a name, then you don't accept to evidence the theorem. Yous can cite the theorem and move on, equally in, 'By the Pythagorean Theorem, \(AC = 3\).' 2. If you are very confident the step is well known but you don't know a name, you can say 'By a well-known theorem, the area of \(ABC\) equals \(rs\), where \(r\) is the inradius and \(s\) is the semiperimeter.' You tin can also leave out the 'By a well-known theorem' scrap, particularly for extremely common results such every bit the 1 just stated. (If you don't know that result, try proving it on your ain.) iii. If you yet aren't sure whether to show a sure stride or assume it's well-known, y'all have a decision to make. If you can evidence it in one or two lines, go ahead and do and so. If it's going to take a lot of piece of work to prove just you know how to do information technology, then at least outline the proof (and give a more thorough one if you have time). If you lot're taking a competition and have no idea how to prove it, cite it and move on. Maybe you'll get lucky and it volition be a 'well-known theorem'. If you lot're writing an educational paper and you don't know how to show information technology, then your paper isn't finished until y'all figure it out. iv. When writing a string of algebraic steps, each step should follow obviously from the one before it. Don't write something like, 'Thus, nosotros have \[ten^2(ten - 4) + {(x +1)}^ii + 5x - 4(x +ii) = -2,\] so \(10=1\) is the but solution.' Yous should include clear elementary steps that arrive clear that the to a higher place is equivalent to \((x - 1)^3 = 0\). 5. You lot can invoke symmetry or illustration when the cases are precisely the same. For example, suppose you want to bear witness that the surface area of any triangle \(ABC\) is given by \[ [ABC] = (ab \sin C + bc \sin A + ac \sin B)/six,\] where \(a = BC\), \(b = Air-conditioning\), and \(c = AB\), and \([ABC]\) is the expanse of \(ABC\). If you lot prove that \[[ABC] = (ab/2)(\sin C),\] then you can only write, 'Similarly, nosotros have \[[ABC] = (bc/2)(\sin A) = (ac/2)(\sin B).'\] 6. When in uncertainty, explain it. Many of the solutions presented in this commodity have a little overkill in them. It'south better to prove too much than besides little.

Here is a sample problem:

Trouble:Find all integral solutions to the equation $$(m^2 + 1)(n^2 + ane) + two(m - north)(i - mn) = 4(mn + 1).$$

(Problem by Titu Andreescu).

This is totally unacceptable:

How Not to Write the Solution ane:

$$(one, 2), (-iii, 0), (0, 3), (-2, -i), (1, 0), (-3, 2), (0, -one), (-2, three)$$

The above is an answer, not a solution. This 'solution' lacks any show that these solutions actually work, and doesn't evidence that at that place are no other solutions. Moreover, it brings the reader no closer to agreement the solution.

How Not to Write the Solution 2:

The given equation rearranges to $$(m + ane)(n - ane) = \pm 2,$$ so the solutions are \[(1, ii), (-3, 0), (0, 3), (-2, -1), (one, 0), (-3, 2), (0, -i), (-2, iii).\]

The to a higher place solution is amend than the first 1; a motivated reader at least has a blink of a path to the solution, but it'southward not at all clear how the original equation rearranges to the given equation, nor how the testify solutions follow.

How to Write the Solution:

How to Write the Solution: We expand the first term and the right-manus side and so regroup terms: $$(one thousand^ii+ i)(n^ii + 1) + two(thou - north)(1 - mn) = 4(mn + 1)$$ $$m^2n^2+ m^2 + n^2+ 1 + 2(m - n)(1 - mn) = 4mn + 4$$ $$one thousand^2n^2- 2mn + i + m^2 - 2mn + n^2 + 2(m - northward)(1 - mn) = four$$ $$(mn - ane)^ii + (m - n)^2 - 2(m - northward)(mn - 1) = 4$$ This left side is the foursquare of \([(mn - 1) - (m - n)]\), so we have: $$[(mn - 1)(m - due north)]^2 = 4$$ $$[mn - one thousand + n - 1]^2 = 4$$ $$[(thou + 1)(northward - 1)]^2 = 4$$ $$(grand + 1)(n - 1) = \pm2,$$ Example 1: For \((m + 1)(due north - 1) = two\), we have the systems of equations: \begin{align*} grand + ane &= 1 \\ northward - ane &= 2 \\ &(0, 3) \end{align*} \begin{align*} m + i &= ii \\ due north - 1 &= one \\ &(1, 2) \end{align*} \begin{align*} grand + 1 &= -i \\ n - 1 &= -two \\ &(-ii, -i) \stop{align*} \brainstorm{align*} m + ane &= -two \\ north - 1 &= -1 \\ &(-3, 0) \end{align*} Case ii: For \((grand + 1)(northward - ane) = -ii\), nosotros have the systems of equations: \begin{align*} m + one &= 1 \\ north - ane &= -two \\ &(0, -1) \end{marshal*} \begin{align*} thousand + 1 &= -two \\ n - 1 &= 1 \\ &(-3, 2) \end{align*} \brainstorm{align*} m + 1 &= -one \\ n - ane &= 2 \\ &(-2, iii) \end{align*} \begin{align*} k + i &= 2 \\ north - 1 &= -1 \\ &(1, 0) \cease{marshal*} Thus, the solutions are \((1, two),\) \((-3, 0),\) \((0, three),\) \((-2, -i),\) \((1, 0),\) \((-iii, two),\) \((0, -1),\) \((-ii, three).\)

Follow the lemmas

Frequently you lot will take to prove multiple preliminary items before tackling the primary problem. In writing a proof, we often choose to dissever these parts from the main proof by labeling each as a 'Lemma' and conspicuously delimiting the lemma and its proof from the rest of the solution.

Hither's a sample problem with two unlike solutions that employ lemmas. We've used a piffling overkill in writing the solutions with lemmas to highlight how well nosotros can clarify solutions with lemmas. Both of these solutions are made significantly easier to read past conspicuously breaking the solution into pieces.

Problem: From vertex \(A\) of triangle \(ABC\), perpendiculars \(AM\) and \(AN\) are fatigued to the bisectors of the exterior angles of the triangle at \(B\) and \(C\). Prove that \(MN\) is equal to half the perimeter of \(ABC\).

Hither are two solutions.

How Not to Write the Solution 1:

Let the line through \(A\) parallel to \(BC\) come across line \(BM\) at \(J\). Permit the line through \(J\) parallel to \(AB\) meet line \(BC\) at \(K\). Let \(MN\) hit \(AB\) at \(10\) and \(Air conditioning\) at \(Y\). Since \(JK \parallel AB\) and \(AJ \parallel BK\), \(JKBA\) is a parallelogram. Since $$ \angle{ABM} = (\angle A + \bending C)/ii.$$ From right triangle \(BAM\), we discover $$\angle BAM = 90 - \angle ABM = \angle B/two.$$ Since \(AJ \parallel BC\), nosotros accept \(\angle JAB = \angle B\), so \(MA\) bisects \(\angle BAJ\). Thus, \(\angle JAM =\) \(\angle BAM\) and triangles \(BAM\) and \(JAM\) are congruent by ASA. Thus, \(AJ = AB\) and parallelogram \(JKBA\) is a rhomb. The diagonals of a rhomb bisect each other, so triangles \(AMJ\) and \(KMB\) are congruent. Thus, the altitudes from \(K\) to \(AJ\) and \(BK\) are equal and \(Thou\) is equidistant from lines \(AJ\) and line \(BK\). By symmetry, this is also true for \(Northward\). Thus, \(MN \parallel BC\) and \(MN\) is equidistant from \(AJ\) and \(BC\). Since \(MX \parallel BK\), \(\triangle AMX \sim \triangle AKB\). Since \(AM =\) \(AK/2\), we take \(MX =\) \(KB/2\). Since \(JKBA\) is a rhomb, \(KB = AB\), so \(MX =\) \(AB/2\), as desired. By symmetry, we likewise have \(NY =\) \(AC/ii\). Since \(XY\) is the midline of \(ABC\) parallel to side \(BC\), we have \(XY =\) \(BC/2\). Thus, $$MN = MX + XY + YN = AB/ii + AC/ii + BC/2,$$ as desired.

(Solution method found past community member fanzha in the Olympiad Geometry course)

Clear Casework

Sometimes the solution to a trouble comes down to investigating a few different cases. In your solution, yous should identify the cases clearly and show that these cases encompass all possibilities.

Hither's a sample trouble:

Problem: How many positive 3-digit integers are such that 1 digit equals the product of the other 2 digits?

(This problem comes from the Fine art of Problem Solving Introduction to Counting & Probability class.)

Here are two solutions:

How Not to Write the Solution:

At that place are the 9 hundreds. There are \(248\), \(284\), \(482\), \(428\), \(824\), and \(842\). There are \(339\), \(933\), and \(393\). There is \(236\), and 5 others just like with \(248\). There are also iii numbers similar \(224\). As well, there are \(111\) and \(122\) and \(133\) and \(144\) and and so on. Each of those tin exist ordered in 3 ways, except for the \(111\), which can be ordered in only one mode. Then, there are 52.

The solution above is short, and the respond is correct, but it's not at all clear that all possibilities have been discovered. Besides, it's pretty tough to see that we have institute exactly 52 solutions – the reader is forced to go through and count themselves.

The solution below clearly covers all possible cases and leaves no incertitude that the total is 52.

How to Write the Solution:

We split up our investigation into cases based on the smallest digit of each number. Case one: The smallest digit is \(0\). If the smallest digit is \(0\), then the number must contain a second \(0\). Thus, this case consists of numbers of the form \(n00\), where \(i \leq n \leq 9\) is whatsoever digit from \(1\) to \(nine\). There are thus 9 numbers with smallest digit \(0\) that satisfy the problem. Case 2: The smallest digit is \(i\). If the smallest digit is \(1\), the number must be of the form \(nn1\), or permutations of this grade (i.eastward. \(n1n\) or \(1nn\)). Still, these 3 permutations are the same when \(due north = ane\). Hence, we take 3 permutations each for \(two \leq n \leq 9\) and but \(ane\) for \(n = 1\), for a total of \(1 +\) \(3(8) =\) \(\mathbf{25}\) numbers with smallest digit \(one\) that satisfy the problem. Case 3: The smallest digit is \(2\). If the smallest digit is \(two\), then the number is of the course \(2mn\), where \(n = 2m\), and permutations of this form. Our merely options here are \((m,n) = (2,4)\), which gives the states iii numbers \((224, 242, 422)\), \((m,due north) = (3,half-dozen)\), which gives us 6 numbers (permutations of \(236\)), and \((m,n) = (4,eight)\), which likewise gives usa 6 numbers. Hence, there are \(3 +\) \(vi +\) \(6 =\) \(\mathbf{15}\) numbers with smallest digit \(2\). Instance 4: The smallest digit is \(3\). At that place are 3 solutions in this case: \(339\), \(393\), \(933\). Case 5: The smallest digit is larger than \(iii\). If the smallest digit is larger than \(3\), the smallest production we can grade with two of the digits is \(iv(4) = 16\), which is not a unmarried digit number. Hence, there are no numbers that satisfy the problem with smallest digit larger than \(3\). Since every possible three-digit number falls in exactly 1 of these cases, we conclude that there are $$9 + 25 + fifteen + iii = 52$$ numbers that satisfy the trouble.

‍

Proofreed

Comunicacating circuitous idas is not ease and can b even harder wen don't edit the presentaion of those ideas for our adience. It pays to oganize are work in ways taht are piece of cake to read to be sur that the audiense gets the point, and to bee sure that your saying what you meen.

If I always wrote that mode, nobody would ever read anything I wrote.

Proof-read and edit your work. God may do crosswords in pen, but you're going to make mistakes. Making certain that you wrote in a way that expresses your ideas clearly and correctly is second in importance only to having the right respond.

Make sure your equations and inequalities apply your variables the way you intend. Yous don't want to write "abc + bcd" when you mean "abd + acd." This not but makes deciphering the residual of your proof hard simply might also throw off your ain calculations.

Exercise writing proofs. Nosotros all make occasional spelling or grammar errors, just the effects of errors multiply and too many of them make otherwise skilful ideas unreadable. Think that "repetition is the mother of all skill."

Problem:

\(ten\), \(y\), and \(z\) are real numbers such that \[10 + y + z = 5 \quad\text{and}\quad xy + yz + zx = three.\] Decide with proof the largest value that whatsoever one of the three numbers tin exist.

Make up one's mind with proof the largest value that whatever one of the three numbers can be.

If all proofs were written this poorly I would cry:

How Not to Write the Solution:

We volition manipulate the given equations to make utilise of the fact that the square of any real number is negative:

(𝑥+𝑦)2=(5–𝑥)2,

𝑥𝑦=three–𝑧(𝑥+𝑦)=3–𝑧(five–𝑧).

At present we note that

0≤(𝑥–𝑦)two=(𝑥+𝑦)–four𝑥𝑦.

We can substitute for both 𝑥+𝑦 and 𝑥𝑦 giving us an inequality involving simply the variable 𝑧:

0≤(𝑥+𝑦)2–4𝑥𝑦=25–10𝑧+𝑧ii–12+20𝑧–4𝑧2=three𝑧ii+10𝑧+13.

Since this inequality holds for 𝑧 nosotros can make up one's mind all possible values of 𝑧:

0≥−three𝑧two+x𝑧+thirteen=−(𝑧+ane)(3𝑧–13).

The iequality holds when −1≥𝑧≥13/iii.

Since the given equations for 𝑥, 𝑦, and 𝑧 can be manipulated to same quadratic inequality in 𝑥, 𝑦, or 𝑧, they each take a minimum of 13/3. This happens when
𝑥=𝑦=13 and 𝑧=thirteen/3.

Graders will be happier when reading this solution:

How to Write the Solution:

Nosotros will manipulate the given equations to brand use of the fact that the square of any real number is nonnegative:

(𝑥+𝑦)2=(5–𝑥)two,

𝑥𝑦=3–𝑧(𝑥+𝑦)=3–𝑧(5–𝑧).

Now we note that

0≤(𝑥–𝑦)2=(𝑥+𝑦)2–4𝑥𝑦.

Nosotros tin substitute for both 𝑥+𝑦 and 𝑥𝑦 giving united states of america an inequality involving but the variable 𝑧:

0≤(𝑥+𝑦)ii–four𝑥𝑦=25–ten𝑧+𝑧2–12+twenty𝑧–4𝑧ii=−iii𝑧ii+10𝑧+13.

Since this inequality holds for z we can determine all possible values of z:

0≤−3𝑧2+10𝑧+thirteen=−(𝑧+1)(3𝑧–thirteen).

The inequality holds when −i≤𝑧≤xiii/3.

Since the given equations for 𝑥, 𝑦, and 𝑧 can exist manipulated to course this aforementioned quadratic inequality in𝑥, 𝑦, or 𝑧, they each have maximum possible values of 13/3. This maximum tin can exist achieved when 𝑥=𝑦=ane/3 and 𝑧=13/3.

‍

Bookends

We have several shelves full of math books in our offices. When we don't have bookends on either end, eventually the books at the ends fall over. Then more autumn over, and so more, and it's a hassle to find and retrieve books without spilling others all over the place.

Similarly, when you have a complicated solution, you should place bookends on your solution so the reader doesn't get lost in the middle. First off proverb what y'all're going to practice, and then do information technology, then say what you lot did. Explaining your general method before doing it is particularly important with standard techniques such every bit contradiction or consecration. For example, you might first with, 'We will show past contradiction that there are infinitely many primes. Assume the reverse, that there are exactly 𝑛 primes ….'

When yous finish your solution, make information technology clear you lot are finished. State the last issue, which should be saying that yous did exactly what the trouble asked yous to do, e.yard. 'Thus, nosotros have shown past contradiction that there are infinitely many prime numbers.' You can also decorate the end of proofs with such items equally 𝑄𝐸𝐷 or 𝐴𝑌𝐷 or 𝑊𝑊𝑊𝑊𝑊 or Undefined control sequence \blacksquare or //.

Here's a sample problem:

Problem: Let 𝐼 be the incenter of triangle 𝐴𝐵𝐶. Testify

(𝐼𝐴)(𝐼𝐵)(𝐼𝐶)=4𝑟2

where 𝑅 is the circumradius of 𝐴𝐵𝐶 and 𝑟 is the inradius of 𝐴𝐵𝐶.

(Problem from Sam Vandervelde of the Mandelbrot Contest . Annotation that the incenter of a triangle is the center of the circle inscribed in the triangle. The inradius is the radius of this circle. The circumradius is the radius of the circle that passes through the vertices of 𝐴𝐵𝐶.)

As this is our last problem, nosotros'll include many of our no-nos in the 'How Not' solution. Good luck piecing it together.

How Not to Write the Solution:

From △𝐴𝐼𝐶 we have ∠𝐴𝐼𝐶= 180∘–∠𝐴𝐶𝐼–∠𝐶𝐴𝐼= 180∘–𝛼/2–𝛾/2= 180∘–(180∘–𝛽)/two= 90∘+𝛽/ii and from △𝐸𝐵𝐶 nosotros have ∠𝐸𝐵𝐶= ∠𝐴𝐵𝐶+∠𝐴𝐵𝑄= 𝛽+(180∘–𝛽)/2= 90∘+𝛽/ii, so ∠𝐴𝐼𝐶=∠𝐸𝐵𝐶. Thus, △𝐴𝐼𝐶∼△𝐸𝐵𝐶 by Angle-Angle Similarity. By symmetry, we conclude △𝐵𝐼𝐶∼△𝐸𝐴𝐶.

[𝐴𝐼𝐸]=(𝐴𝐼)(𝐴𝐸)/2= (𝑥)(𝑏𝑦/𝑧)/2=𝑏𝑥𝑦/2𝑧, [𝐴𝐼𝐶]=𝑏𝑟/2, and [𝐸𝐵𝐶]=[𝐴𝐼𝐶](𝐵𝐶/𝐼𝐶)2= (𝑏𝑟/ii)(𝑎/𝑧)2= 𝑎2𝑏𝑟/2𝑧2, so [𝐸𝐴𝐶𝐵]= 𝑏𝑥𝑦/2𝑧+𝑏𝑟/2+ 𝑎2𝑏𝑟/2𝑧ii= 𝑎𝑥𝑦/2𝑧+𝑎𝑟/2+ 𝑎𝑏2𝑟/ii𝑧ii. Thus, (𝑏–𝑎)(𝑥𝑦/ii𝑧+𝑟+𝑎𝑏𝑟/2𝑧2)=0. If 𝑏=𝑎, then 𝑎𝑏–𝑧2=𝑥𝑦𝑧/𝑟 follows from the Pythagorean Theorem and the Bending Bisector Theorem. Otherwise, 𝑎𝑏–𝑧2=𝑥𝑦𝑧/𝑟 follows immediately.

Draw altitude 𝐼𝐸 of 𝐴𝐼𝐶. [𝐸𝐼𝐶]= (𝑧2/ii)sin𝛾= 𝑟(𝑠–𝑐)/two, and [𝐴𝐵𝐶]= (𝑎𝑏/ii)sin𝛾= 𝑟𝑠, so [(𝑎𝑏–𝑧2)/two]sin𝛾=𝑟𝑐. Then the Law of Sines and the earlier equation requite our result.

Short, ugly, and completely incomprehensible.

How to Write the Solution:

We let

𝑎=𝐵𝐶,𝑏=𝐴𝐶,𝑐=𝐴𝐵

𝑠=(𝑎+𝑏+𝑐)/2

𝛼=∠𝐵𝐴𝐶,𝛽=∠𝐴𝐵𝐶,𝛾=∠𝐴𝐵𝐶,

[𝐴𝐵𝐶]=area of polygon 𝐴𝐵𝐶

𝑥=𝐼𝐴,𝑦=𝐼𝐵,𝑧=𝐼𝐶

Let the external bisectors of angles 𝐴 and 𝐵 of △𝐴𝐵𝐶 meet at 𝐸 equally shown. Point 𝐸 is equidistant from lines 𝐴𝐵, 𝐴𝐶, and 𝐵𝐶, and then information technology is on angle bisector 𝐶𝐼 besides. Nosotros volition show

[𝐸𝐴𝐶𝐵]=𝑏𝑥𝑦/two𝑧+𝑏𝑟/two+𝑎2𝑏𝑟/2𝑧2=𝑎𝑥𝑦/2𝑧+𝑎𝑟/2+𝑎𝑏2𝑟/2𝑧2²,(1)

and

(sin𝛾)(𝑎𝑏–𝑧2)/2=𝑟𝑐.(two)

From (1) we will show that 𝑎𝑏–𝑧2= 𝑥𝑦𝑧/𝑟, which we will combine with (2) and known triangle relationships to evidence the desired effect.

Lemma 1: △𝐴𝐼𝐶∼△𝐸𝐵𝐶 and △𝐵𝐼𝐶∼△𝐸𝐴𝐶.
Proof: Past symmetry, the two results are equivalent. We will show the first. Since 𝐶𝐼 bisects ∠𝐴𝐶𝐵, nosotros accept ∠𝐴𝐶𝐼=∠𝐵𝐶𝐸.

From △𝐴𝐼𝐶 nosotros take

∠𝐴𝐼𝐶=180∘–∠𝐴𝐶𝐼–∠𝐶𝐴𝐼=180∘–𝛼/2–𝛾/ii=180∘–(180∘–𝛽)/2=90∘+𝛽/2

and from △𝐸𝐵𝐶 nosotros have

∠𝐸𝐵𝐶=∠𝐴𝐵𝐶+∠𝐴𝐵𝑄=𝛽+(180∘–𝛽)/2=90∘+𝛽/2,

so △𝐴𝐼𝐶∼△𝐸𝐵𝐶 past Bending-Bending Similarity. By symmetry, we conclude △𝐴𝐼𝐶∼△𝐸𝐵𝐶. Undefined control sequence \blacksquare

Lemma 2:

[𝐸𝐴𝐶𝐵]=𝑏𝑥𝑦/2𝑧+𝑏𝑟/two+𝑎two𝑏𝑟/2𝑧two=𝑎𝑥𝑦/ii𝑧+𝑎𝑟/2+𝑎𝑏2𝑟/two𝑧two

Proof: We detect the surface area of 𝐸𝐴𝐶𝐵 by splitting it into pieces:

[𝐸𝐴𝐶𝐵]=[𝐴𝐼𝐸]+[𝐴𝐼𝐶]+[𝐸𝐵𝐶]

Offset nosotros tackle [𝐴𝐼𝐸] by showing it is a right triangle with legs 𝑥 and 𝑏𝑦/𝑧. From Lemma 1, we accept △𝐵𝐼𝐶∼△𝐸𝐴𝐶. Hence, 𝐴𝐸/𝐼𝐵=𝐴𝐶/𝐼𝐶, or

𝐴𝐸=(𝐴𝐶)(𝐼𝐵)/𝐼𝐶=𝑏𝑦/𝑧.

Since

[𝐴𝐼𝐸]=(𝐴𝐼)(𝐴𝐸)/2=(𝑥)(𝑏𝑦/𝑧)/2=𝑏𝑥𝑦/2𝑧.(iii)

For triangle 𝐴𝐼𝐶 nosotros note that the altitude from 𝐼 to 𝐴𝐶 is the inradius of 𝐴𝐵𝐶, so

[𝐴𝐼𝐶]=𝑏𝑟/2.(four)

Finally, since △𝐴𝐼𝐶∼△𝐸𝐵𝐶, nosotros accept

[𝐸𝐵𝐶]=[𝐴𝐼𝐶](𝐵𝐶/𝐼𝐶)ii=(𝑏𝑟/2)(𝑎/𝑧)2=𝑎ii𝑏𝑟/2𝑧2.(five)

Calculation (3), (4), and (5) yields

[𝐸𝐴𝐶𝐵]=𝑏𝑥𝑦/2𝑧+𝑏𝑟/2+𝑎two𝑏𝑟/two𝑧2(half dozen)

By symmetry, we note that [𝐸𝐴𝐶𝐵] too equals our expression in (vi) with 𝑎 and 𝑏 interchanged and 𝑥 and 𝑦 interchanged. Hence, we have the desired

[𝐸𝐴𝐶𝐵]=𝑏𝑥𝑦/2𝑧+𝑏𝑟/two+𝑎2𝑏𝑟/2𝑧2=𝑎𝑥𝑦/2𝑧+𝑎𝑟/two+𝑎𝑏two𝑟/2𝑧2

Undefined control sequence \blacksquare

Lemma 3 : 𝑎𝑏–𝑧2=𝑥𝑦𝑧/𝑟.
Proof: Rearranging our result from Lemma 1 yields

(𝑏𝑥𝑦/2𝑧+𝑏𝑟/2+𝑎2𝑏𝑟/ii𝑧2)–(𝑎𝑥𝑦/ii𝑧+𝑎𝑟/2+𝑎𝑏two𝑟/two𝑧2)=0

(𝑏𝑥𝑦/two𝑧–𝑎𝑥𝑦/2𝑧)+(𝑏𝑟/two–𝑎𝑟/ii)+(𝑎ii𝑏𝑟/2𝑧2–𝑎𝑏2𝑟/two𝑧2)=0

(𝑏−𝑎)(𝑥𝑦/2𝑧)+(𝑏–𝑎)(𝑟/2)–(𝑏–𝑎)(𝑎𝑏𝑟/ii𝑧2)=0

(𝑏–𝑎)(𝑥𝑦/two𝑧+𝑟/two–𝑎𝑏𝑟/2𝑧2)=0

Thus, one of the terms in this product equals 0.

Case one: 𝑏–𝑎=0.

If 𝑏=𝑎 then 𝐴𝐵𝐶 is isosceles and 𝛼=𝛽. Hence, the extension of angle bisector 𝐶𝐼 is perpendicular to 𝐴𝐵 at signal 𝐷 as shown. Since 𝐼 is the incenter of 𝐴𝐵𝐶 and 𝐼𝐷⊥𝐴𝐵, 𝐼𝐷=𝑟 since 𝐼𝐷 is an inradius of 𝐴𝐵𝐶. As well,

∠𝐼𝐴𝐵=𝛼/2=𝛽/2=∠𝐼𝐵𝐴,

so 𝐼𝐵=𝐼𝐴 (i.e. 𝑥=𝑦). Thus, the equation we wish to prove, 𝑎𝑏–𝑧two= 𝑥𝑦𝑧/𝑟, is in this case equivalent to

𝑎two–𝑧two=𝑥2𝑧/𝑟.(7)

From correct triangles 𝐶𝐴𝐷 and 𝐼𝐴𝐷, we have

(𝑐/two)2+𝑟2=𝑥two,(8)

(𝑐/ii)ii+(𝑧+𝑟)2=𝑎2.(9)

The Bending Bisector Theorem gives us 𝑎/𝑧= 𝐴𝐶/𝐶𝐼= 𝐴𝐷/𝐷𝐼= (𝑐/2)/𝑟, or

𝑐/2=𝑎𝑟/𝑧.(10)

Substituting (10) into (8) yields

𝑎2𝑟2/𝑧2+𝑟two=𝑥2,

𝑎2𝑟ii+𝑟two𝑧2=𝑥two𝑧2,

(𝑟/𝑧)(𝑎2+𝑧2)=𝑥ii𝑧/𝑟.(11)

Substituting (ten) into (9) gives

𝑎2𝑟two/𝑧two+(𝑧+𝑟)2=𝑎2,

𝑎ii𝑟2+𝑧two(𝑧+𝑟)ii=𝑎ii𝑧2,

𝑧two(𝑧+𝑟)2=𝑎ii𝑧2–𝑎2𝑟2,

𝑧2(𝑧+𝑟)ii=𝑎ii(𝑧ii–𝑟2),

𝑧(𝑧+𝑟)=𝑎ii(𝑧–𝑟),

𝑎2𝑟+𝑧2𝑟=𝑎ii𝑧–𝑧3,

(𝑟/𝑧)(𝑎two+𝑧ii)=𝑎two–𝑧2.(12)

Combining (xi) and (12) gives usa the desired 𝑎2–𝑧2=𝑥two𝑧/𝑟.

Case 2: 𝑥𝑦/two𝑧+𝑟/2–𝑎𝑏𝑟/2𝑧2=0.

Multiplying this equation by 2𝑧2/𝑟 yields

𝑥𝑦𝑧/𝑟+𝑧2–𝑎𝑏=0,

from which the desired 𝑎𝑏–𝑧2=𝑥𝑦𝑧/𝑟 immediately follows.

Thus, the lemma is proved. Undefined control sequence \blacksquare

Lemma 4: (sin𝛾)(𝑎𝑏–𝑧2)/2=𝑟𝑐.

Proof:

We draw distance 𝐼𝐹 perpendicular to 𝐵𝐶 as shown. Nosotros employ the following known triangle relationships:

𝐶𝐹[𝐴𝐵𝐶]=𝑠–𝑐=(𝑎𝑏/2)sin𝛾=𝑟𝑠

Just every bit [𝐴𝐵𝐶]=(𝑎𝑏/2)sin𝛾, we accept

𝐶𝐹𝐶𝐹𝐶𝐹𝐶𝐹=[(𝐶𝐹)(𝐼𝐶)/ii]sin(𝛾/two)=[𝑧cos(𝛾/2)][𝑧/2sin(𝛾/two)]=(𝑧2/2)cos(𝛾/2)sin(𝛾/two)=(𝑧2/4)sin𝛾

where nosotros accept used sin2𝛾= 2sin𝛾cos𝛾 in the last step.

Since 𝐶𝐹𝐼 is right, we take [𝐶𝐹𝐼]=𝑟(𝑠–𝑐)/2. Hence, nosotros take two expressions for [𝐴𝐵𝐶]–2[𝐶𝐹𝐼]:

(𝑎𝑏/2)sin𝛾–2(𝑧2/4)sin𝛾[(𝑎𝑏–𝑧ii/2]sin𝛾=𝑟𝑠–2[𝑟(𝑠–𝑐/two]=𝑟𝑠

as desired. Undefined control sequence \blacksquare

Nosotros now complete our proof. Dividing the issue of Lemma iv by (sin𝛾)/2 gives

𝑎𝑏–𝑧2=2𝑟𝑐/(sin𝛾).

Since Lemma 3 gives the states 𝑎𝑏–𝑧two=𝑥𝑦𝑧/𝑟 and the Extended Law of Sines gives u.s.a. sin𝛾=𝑐/ii𝑅, the equation in a higher place becomes

𝑥𝑦𝑧/𝑟𝑥𝑦𝑧/𝑟𝑥𝑦𝑧=2𝑟𝑐/(𝑐/2𝑅)=iv𝑅𝑟=4𝑅𝑟ii

Thus, we have shown that if 𝐼 is the incenter of triangle 𝐴𝐵𝐶, nosotros have

(𝐼𝐴)(𝐼𝐵)(𝐼𝐶)=iv𝑅𝑟2,

where 𝑅 is the circumradius of 𝐴𝐵𝐶 and 𝑟 is the inradius of 𝐴𝐵𝐶.

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Notation, we proved some intermediate results we probably didn't have to (such every bit the fact that 𝐸 is on ray 𝐶𝐼) when the results were quick and easy to prove. Others we stated by fiat, such every bit [𝐴𝐵𝐶]=𝑟𝑠, since the proofs are more involved, and nosotros feel pretty prophylactic that these results tin can exist cited as known results without proof.

The above is a pretty daunting proof. What our solution doesn't give is any indication of how we might have come up upward with this solution. If y'all didn't find the above solution on your own, see if y'all can figure out how you might have come up with it now that you have seen it.

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Source: https://artofproblemsolving.com/news/articles/how-to-write-a-solution

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