ouβve figured out the solution to the problem β fantastic! But youβre not finished. Whether you are writing solutions for a competition, a journal, a message board, or just to show off for your friends, you must master the art of communicating your solution clearly.

Brilliant ideas and innovative solutions to problems are pretty worthless if you canβt communicate them. In this article, we explore many aspects of how to write a clear solution. Below is an index; each page of the article includes a sample βHow Not Toβ solution and βHow Toβ solution. One common theme youβll find throughout each point is that every time you make an experienced reader have to think to follow your solution, you lose.

As you read the βHow Toβ solutions, you may think some of them are overwritten. Indeed, some of them could be condensed. Some steps we chose to prove could probably be cited without proof. However, it is far better to prove too much too clearly than to prove too little. Rarely will a reader complain that a solution is too easy to understand or too easy on the eye.

One note of warning: many of the problems we use for examples are extremely challenging problems. Beginners, and even intermediate students, should not be upset if they have difficulty solving the problems on their own.

### Table of Contents:

- Have a Plan
- Readers Are Not Interpreters
- U s e S p a c e
- sdrawkcaB knihT, Write Forwards
- Name Your Characters
- A Picture is Worth a Thousand Words
- Solution Readers, Not Mind Readers
- Follow the Lemmas
- Clear Casework
- Proofreed
- Bookends

β

Your goal in writing a clear solution is to prevent the reader from having to think. You must express your ideas clearly and concisely. The experienced reader should never have to wonder where you are headed, or why any claim you make is true. The first step in writing a clear solution is having a plan. Make a simple outline of your solution. Include the items youβll need to define, and the order in which you will write up the important parts of your solution. The outline will help ensure that you donβt skip anything and that you put your steps in an order thatβs easy to follow.

**Hereβs a sample problem:**

A sphere of radius π 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 π, π, π, π. Show that:

π+π+π+π=2π

Hereβs a solution that looks short but is pretty tough to read.

**How Not to Write the Solution:**

*(General solution method found by community member **zabelman** in the Olympiad Geometry class.)*

The main problem with the above solution is one of organization. We defined variables after they popped up. Midway through the solution we sidetracked to prove the volume of ABCD is ππ/3. Sometimes we wrote important equations right in our paragraphs instead of highlighting them by giving them their own lines.

If we outline before writing the solution, we wonβt have these problems. We can list what we need to define, decide what items we need to prove before our main proof (we 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 define: π΄π΅πΆπ·,βπ,π,[π΄],π΄πππ.

Order of things to prove:

β

This list looks obvious once you have it written up, but if you just plow ahead with the solution without planning, you may end up skipping items and having to wedge them in as we did in our βHow Not to Write the Solutionβ.

**How to Write the Solution:**

*(General solution method found by community member **zabelman** in the Olympiad Geometry class.)*

β

The first thing a reader sees on your paper isnβt the structure of your solution. It isnβt the answer, it isnβt the words you choose. Itβs how the solution sits on the paper. If the reader has to decipher scrawl, youβre going to lose him. Ideally, youβll typeset your solution with a program like LaTeX. However, in most contests you donβt have the luxury of turning to a computer and youβll have to write it out by hand. There are few very important rules of thumb when writing a solution by hand. Many are obvious, some are less so. You should follow them all.

- Use blank paper. Donβt use graph paper or lined paper β the lines often make solutions harder to read. Never use paper that is torn out of a spiral notebook.
- Respect margins. If you are starting with a completely blank piece of paper, draw the margins on all four sides (top, bottom, right, left). Make your margins at least 0.5 inches, and preferably a full inch.
- Write horizontally; never turn your writing when you reach the end of a line in order to jam in a little more information. You can always start a new line or a new page.
- Leave space at the top for a βPage _ of _β so the reader knows how many pages there are, and what page sheβs on. You probably wonβt know how many pages youβll write when you start, but you can fill these out when youβre finished. If you get to the bottom of a page and your solution must continue on another page, write βContinuedβ at the bottom of that page so the reader knows weβre not finished. (This also helps readers know if theyβre missing pages.)
- Donβt write in cursive. Print. And print clearly.
- Use pen. If you must use pencil, do not erase β the smudges from erasers make a mess.
- When you make a mistake youβd like to omit, draw a single line through it and move on. If itβs a large block to omit, draw an βXβ through it and move on. Donβt scribble out large blocks of text.
- If you left something out and want to add it at the end, put a simple symbol, like a (*), at the point where you would like the new text to be considered added, and leave a brief note, such as βProof below.β Below, you can write β(*) Addendum:β and proceed with the proof. Donβt use a bunch of arrows to direct the reader all over the page.

Below are two solutions. Neither solution is picture-perfect; when youβre under time pressure, itβs hard to write perfect-looking proofs. You should find the second one much more enjoyable to read. When youβre writing solutions, keep the above tips in mind, and just remember, βIf they canβt read it, itβs not right.β

**How Not 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:**

β

imagineyoutrytoreadaparagraphoftextthathasnopuncuationnocapitalsandjustenough

spaceinittobreakuplinessoitdoesntmessupbrowsersitsreallytoughtoreadandpretty

soonyoulldecidethatitsnotworthreadingandyoullgoandreadsomethingelseyouwont

realizehowterriblyharditistotypelikethisitshardbecausewhenyoureusedtowritingcle

arlyandusingspaceandpunctuationandsentencestructureitgetsreallyhardtowritewith

outitsimilarlyonceyougetusedtoproperlyusingspaceinwritingyoursolutionsitwillbesec

ondnatureandyoullactuallyfindithardtowriteanindecipherableproof

When you write your solution you should:

- Give each important definition or equation its own line.
- Donβt bury too much algebra in a paragraph. You can write line after line of algebra, but put each step on its own line. Donβt cram the algebra in a paragraph.
- Label equations or formulas or lemmas or cases you will use later very clearly.
- Remember that thereβs always more paper.

β

Have fun reading this solution.

**How Not to Write the Solution:**

Hereβs the same solution, with nearly the same wording.

**How to Write the Solution:**

Which would you rather read?

β

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

βFiguring out how to make an omelette is easy. Anybody who has eaten an omelette knows that an omelette is typically made 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 end up inside the egg means that we should begin cooking the eggs flatly on a pan and then add the ingredients. We can then roll part of the egg over the ingredients so as to trap them on the inside. If we needed some of the ingredients precooked we could do that before adding them to the eggsβ¦β

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

βPrepare vegetables and other desired omelette fillings. Beat eggs. Start cooking the eggs. Add your fillings in the middle so that part of the egg can be pulled over the ingredients. When the omelette is closed, continue to cook and flip the omelette until the eggs look well-cooked.β

The reader doesnβt care how the process of cooking an omelette was unraveled by the author. The reader just wants to know how to make an omelette.

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

Hereβs a sample problem.

This solution might be a good way to see how we might come up with a solution from scratch, but itβs not a particularly well-written proof:

**How Not to Write the Solution:**

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

**How to Write the Solution:**

β

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 land and all the men of the great man who lived in a large castle that ruled over the people in the land couldnβt put the large thin-shelled vehicle for a young fowl that was created by a huge female 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 audience will understand and enjoy. Instead of writing about βA large thin-shelled vehicle for a young fowl that was created by a huge female bird,β we call that big egg βHumpty-Dumptyβ and tell the story. Likewise, a well-written proof often involves naming the important quantities or ideas that play a part in the story of your solution. Naming your characters can also help you find solutions to problems, so itβs not something you should wait until proof-writing time to do.

When you do name your characters, you name them simply, clearly, and write up front, so the reader knows exactly where to go to find out exactly who this π person is and what that π(π₯) function stands for.

Hereβs an example problem.

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

**How Not to Write the Solution:**

The solution below is easy to read because the main characters have names. Specifically, we 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 allow the writer to describe the characters more completely and succinctly.

**How to Write the Solution:**

When youβre writing a solution to a geometry problem, or any problem involving a picture, you should include the diagram. If you donβt include the diagram, you often make the grader have to draw it for you. Even if the diagram is given in the problem, you should include it 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. Use a geometry rendering program if you are typesetting your solution, or use a ruler and compass if you are writing your solution by hand.

Hereβs an example.

Hereβs a solution without a diagram.

**How to Write the Solution:**

*(Solution method found by community member **3cnfsat** in the Olympiad Geometry class)*

Hereβs a solution that includes the diagram:

**How to Write the Solution:**

*(Solution method found by community member ***3cnfsat*** in the Olympiad Geometry class)*

If you arenβt familiar with this fact, try to prove it yourself (and write a nice solution). Every good geometer reaches for this fact as easily as they reach for the Pythagorean Theorem.

A full solution does not just mean a correct answer. You should justify every notable step of your solution. An experienced reader should never wonder βWhy is that true?β while reading your solution. She should also never be left in doubt as to whether or not you know why it is true.

Itβs not always clear what steps you can assume the reader understands and what steps you have to explain. Here are a few guidelines:

β

Here is a sample problem:

(Problem by **Titu Andreescu**).

This is totally unacceptable:

**How Not to Write the Solution 1:**

The above is an answer, not a solution. This βsolutionβ lacks any evidence that these solutions actually work, and doesnβt show that there are no other solutions. Moreover, it brings the reader no closer to understanding the solution.

**How Not to Write the Solution 2:**

The above solution is better than the first one; a motivated reader at least has a glimmer of a path to the solution, but itβs not at all clear how the original equation rearranges to the given equation, nor how the show solutions follow.

**How to Write the Solution:**

Often you will have to prove multiple preliminary items before tackling the main problem. In writing a proof, we often choose to separate these parts from the main proof by labeling each as a βLemmaβ and clearly delimiting the lemma and its proof from the rest of the solution.

Hereβs a sample problem with two different solutions that employ lemmas. Weβve used a little overkill in writing the solutions with lemmas to highlight how well we can clarify solutions with lemmas. Both of these solutions are made significantly easier to read by clearly breaking the solution into pieces.

Here are two solutions.

**How Not to Write the Solution 1:**

*(Solution method found by community member **fanzha** in the Olympiad Geometry class)*

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

Hereβs a sample problem:

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

(This problem comes from the Art of Problem Solving Introduction to Counting & Probability course.)

Here are two solutions:

**How Not to Write the Solution:**

The solution above is short, and the answer is correct, but itβs not at all clear that all possibilities have been discovered. Also, itβs pretty tough to see that we have found exactly 52 solutions β the reader is forced to go through and count themselves.

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

**How to Write the Solution:**

β

Comunicacating complex idas is not ease and can b even harder wen donβt edit the presentaion of those ideas for our adience. It pays to oganize are work in ways taht are easy 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 way, nobody would ever read anything I wrote.

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

Make sure your equations and inequalities use your variables the way you intend. You donβt want to write βabc + bcdβ when you mean βabd + acd.β This not only makes deciphering the rest of your proof difficult but might also throw off your own calculations.

Practice writing proofs. We all make occasional spelling or grammar errors, but the effects of errors multiply and too many of them make otherwise good ideas unreadable. Remember that βrepetition is the mother of all skill.β

**Problem:**

If all proofs were written this poorly I would cry:

**How Not to Write the Solution:**

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

(π₯+π¦)2=(5βπ₯)2,

π₯π¦=3βπ§(π₯+π¦)=3βπ§(5βπ§).

Now we note that

0β€(π₯βπ¦)2=(π₯+π¦)β4π₯π¦.

We can substitute for both π₯+π¦ and π₯π¦ giving us an inequality involving only the variable π§:

0β€(π₯+π¦)2β4π₯π¦=25β10π§+π§2β12+20π§β4π§2=3π§2+10π§+13.

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

0β₯β3π§2+10π§+13=β(π§+1)(3π§β13).

The iequality holds when β1β₯π§β₯13/3.

Since the given equations for π₯, π¦, and π§ can be manipulated to same quadratic inequality in π₯, π¦, or π§, they each have a minimum of 13/3. This happens when

π₯=π¦=13 and π§=13/3.

Graders will be happier when reading this solution:

**How to Write the Solution:**

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

(π₯+π¦)2=(5βπ₯)2,

π₯π¦=3βπ§(π₯+π¦)=3βπ§(5βπ§).

Now we note that

0β€(π₯βπ¦)2=(π₯+π¦)2β4π₯π¦.

We can substitute for both π₯+π¦ and π₯π¦ giving us an inequality involving only the variable π§:

0β€(π₯+π¦)2β4π₯π¦=25β10π§+π§2β12+20π§β4π§2=β3π§2+10π§+13.

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

0β€β3π§2+10π§+13=β(π§+1)(3π§β13).

The inequality holds when β1β€π§β€13/3.

Since the given equations for π₯, π¦, and π§ can be manipulated to form this same quadratic inequality inπ₯, π¦, or π§, they each have maximum possible values of 13/3. This maximum can be achieved when π₯=π¦=1/3 and π§=13/3.

β

We have several shelves full of math books in our offices. When we donβt have bookends on either end, eventually the books at the ends fall over. Then more fall over, then more, and itβs 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 doesnβt get lost in the middle. Start off saying what youβre going to do, then do it, then say what you did. Explaining your general method before doing it is particularly important with standard techniques such as contradiction or induction. For example, you might start with, βWe will show by contradiction that there are infinitely many primes. Assume the opposite, that there are exactly π primes β¦.β

When you finish your solution, make it clear you are finished. State the final result, which should be saying that you did exactly what the problem asked you to do, e.g. βThus, we have shown by contradiction that there are infinitely many prime numbers.β You can also decorate the end of proofs with such items as ππΈπ· or π΄ππ· or πππππ or Undefined control sequence \blacksquare or //.

Hereβs a sample problem:

**Problem:** Let πΌ be the incenter of triangle π΄π΅πΆ. Prove

(πΌπ΄)(πΌπ΅)(πΌπΆ)=4π2

where π is the circumradius of π΄π΅πΆ and π is the inradius of π΄π΅πΆ.

*(Problem from **Sam Vandervelde** of the **Mandelbrot Competition**. Note 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 π΄π΅πΆ.)*

As this is our last problem, weβll include many of our no-nos in the βHow Notβ solution. Good luck piecing it together.

**How Not to Write the Solution:**

From β³π΄πΌπΆ we have β π΄πΌπΆ= 180βββ π΄πΆπΌββ πΆπ΄πΌ= 180ββπΌ/2βπΎ/2= 180ββ(180ββπ½)/2= 90β+π½/2 and from β³πΈπ΅πΆ we have β πΈπ΅πΆ= β π΄π΅πΆ+β π΄π΅π= π½+(180ββπ½)/2= 90β+π½/2, so β π΄πΌπΆ=β πΈπ΅πΆ. Thus, β³π΄πΌπΆβΌβ³πΈπ΅πΆ by Angle-Angle Similarity. By symmetry, we conclude β³π΅πΌπΆβΌβ³πΈπ΄πΆ.

[π΄πΌπΈ]=(π΄πΌ)(π΄πΈ)/2= (π₯)(ππ¦/π§)/2=ππ₯π¦/2π§, [π΄πΌπΆ]=ππ/2, and [πΈπ΅πΆ]=[π΄πΌπΆ](π΅πΆ/πΌπΆ)2= (ππ/2)(π/π§)2= π2ππ/2π§2, so [πΈπ΄πΆπ΅]= ππ₯π¦/2π§+ππ/2+ π2ππ/2π§2= ππ₯π¦/2π§+ππ/2+ ππ2π/2π§2. Thus, (πβπ)(π₯π¦/2π§+π+πππ/2π§2)=0. If π=π, then ππβπ§2=π₯π¦π§/π follows from the Pythagorean Theorem and the Angle Bisector Theorem. Otherwise, ππβπ§2=π₯π¦π§/π follows immediately.

Draw altitude πΌπΈ of π΄πΌπΆ. [πΈπΌπΆ]= (π§2/2)sinπΎ= π(π βπ)/2, and [π΄π΅πΆ]= (ππ/2)sinπΎ= ππ , so [(ππβπ§2)/2]sinπΎ=ππ. Then the Law of Sines and the earlier equation give our result.

Short, ugly, and completely incomprehensible.

**How to Write the Solution:**

We let

π=π΅πΆ,π=π΄πΆ,π=π΄π΅

π =(π+π+π)/2

πΌ=β π΅π΄πΆ,π½=β π΄π΅πΆ,πΎ=β π΄π΅πΆ,

[π΄π΅πΆ]=area of polygon π΄π΅πΆ

π₯=πΌπ΄,π¦=πΌπ΅,π§=πΌπΆ

Let the external bisectors of angles π΄ and π΅ of β³π΄π΅πΆ meet at πΈ as shown. Point πΈ is equidistant from lines π΄π΅, π΄πΆ, and π΅πΆ, so it is on angle bisector πΆπΌ as well. We will show

[πΈπ΄πΆπ΅]=ππ₯π¦/2π§+ππ/2+π2ππ/2π§2=ππ₯π¦/2π§+ππ/2+ππ2π/2π§2Β²,(1)

and

(sinπΎ)(ππβπ§2)/2=ππ.(2)

From (1) we will show that ππβπ§2= π₯π¦π§/π, which we will combine with (2) and known triangle relationships to show the desired result.

**Lemma 1**: β³π΄πΌπΆβΌβ³πΈπ΅πΆ and β³π΅πΌπΆβΌβ³πΈπ΄πΆ.**Proof**: By symmetry, the two results are equivalent. We will show the first. Since πΆπΌ bisects β π΄πΆπ΅, we have β π΄πΆπΌ=β π΅πΆπΈ.

From β³π΄πΌπΆ we have

β π΄πΌπΆ=180βββ π΄πΆπΌββ πΆπ΄πΌ=180ββπΌ/2βπΎ/2=180ββ(180ββπ½)/2=90β+π½/2

and from β³πΈπ΅πΆ we have

β πΈπ΅πΆ=β π΄π΅πΆ+β π΄π΅π=π½+(180ββπ½)/2=90β+π½/2,

so β³π΄πΌπΆβΌβ³πΈπ΅πΆ by Angle-Angle Similarity. By symmetry, we conclude β³π΄πΌπΆβΌβ³πΈπ΅πΆ. Undefined control sequence \blacksquare

**Lemma 2**:

[πΈπ΄πΆπ΅]=ππ₯π¦/2π§+ππ/2+π2ππ/2π§2=ππ₯π¦/2π§+ππ/2+ππ2π/2π§2

**Proof**: We find the area of πΈπ΄πΆπ΅ by splitting it into pieces:

[πΈπ΄πΆπ΅]=[π΄πΌπΈ]+[π΄πΌπΆ]+[πΈπ΅πΆ]

First we tackle [π΄πΌπΈ] by showing it is a right triangle with legs π₯ and ππ¦/π§. From Lemma 1, we have β³π΅πΌπΆβΌβ³πΈπ΄πΆ. Hence, π΄πΈ/πΌπ΅=π΄πΆ/πΌπΆ, or

π΄πΈ=(π΄πΆ)(πΌπ΅)/πΌπΆ=ππ¦/π§.

Since

[π΄πΌπΈ]=(π΄πΌ)(π΄πΈ)/2=(π₯)(ππ¦/π§)/2=ππ₯π¦/2π§.(3)

For triangle π΄πΌπΆ we note that the altitude from πΌ to π΄πΆ is the inradius of π΄π΅πΆ, so

[π΄πΌπΆ]=ππ/2.(4)

Finally, since β³π΄πΌπΆβΌβ³πΈπ΅πΆ, we have

[πΈπ΅πΆ]=[π΄πΌπΆ](π΅πΆ/πΌπΆ)2=(ππ/2)(π/π§)2=π2ππ/2π§2.(5)

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

[πΈπ΄πΆπ΅]=ππ₯π¦/2π§+ππ/2+π2ππ/2π§2(6)

By symmetry, we note that [πΈπ΄πΆπ΅] also equals our expression in (6) with π and π interchanged and π₯ and π¦ interchanged. Hence, we have the desired

[πΈπ΄πΆπ΅]=ππ₯π¦/2π§+ππ/2+π2ππ/2π§2=ππ₯π¦/2π§+ππ/2+ππ2π/2π§2

Undefined control sequence \blacksquare

**Lemma 3 **: ππβπ§2=π₯π¦π§/π.**Proof**: Rearranging our result from Lemma 1 yields

(ππ₯π¦/2π§+ππ/2+π2ππ/2π§2)β(ππ₯π¦/2π§+ππ/2+ππ2π/2π§2)=0

(ππ₯π¦/2π§βππ₯π¦/2π§)+(ππ/2βππ/2)+(π2ππ/2π§2βππ2π/2π§2)=0

(πβπ)(π₯π¦/2π§)+(πβπ)(π/2)β(πβπ)(πππ/2π§2)=0

(πβπ)(π₯π¦/2π§+π/2βπππ/2π§2)=0

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

**Case 1**: πβπ=0.

If π=π then π΄π΅πΆ is isosceles and πΌ=π½. Hence, the extension of angle bisector πΆπΌ is perpendicular to π΄π΅ at point π· as shown. Since πΌ is the incenter of π΄π΅πΆ and πΌπ·β₯π΄π΅, πΌπ·=π since πΌπ· is an inradius of π΄π΅πΆ. Also,

β πΌπ΄π΅=πΌ/2=π½/2=β πΌπ΅π΄,

so πΌπ΅=πΌπ΄ (i.e. π₯=π¦). Thus, the equation we wish to prove, ππβπ§2= π₯π¦π§/π, is in this case equivalent to

π2βπ§2=π₯2π§/π.(7)

From right triangles πΆπ΄π· and πΌπ΄π·, we have

(π/2)2+π2=π₯2,(8)

(π/2)2+(π§+π)2=π2.(9)

The Angle Bisector Theorem gives us π/π§= π΄πΆ/πΆπΌ= π΄π·/π·πΌ= (π/2)/π, or

π/2=ππ/π§.(10)

Substituting (10) into (8) yields

π2π2/π§2+π2=π₯2,

π2π2+π2π§2=π₯2π§2,

(π/π§)(π2+π§2)=π₯2π§/π.(11)

Substituting (10) into (9) gives

π2π2/π§2+(π§+π)2=π2,

π2π2+π§2(π§+π)2=π2π§2,

π§2(π§+π)2=π2π§2βπ2π2,

π§2(π§+π)2=π2(π§2βπ2),

π§(π§+π)=π2(π§βπ),

π2π+π§2π=π2π§βπ§3,

(π/π§)(π2+π§2)=π2βπ§2.(12)

Combining (11) and (12) gives us the desired π2βπ§2=π₯2π§/π.

**Case 2**: π₯π¦/2π§+π/2βπππ/2π§2=0.

Multiplying this equation by 2π§2/π yields

π₯π¦π§/π+π§2βππ=0,

from which the desired ππβπ§2=π₯π¦π§/π immediately follows.

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

**Lemma 4**: (sinπΎ)(ππβπ§2)/2=ππ.

**Proof**:

We draw altitude πΌπΉ perpendicular to π΅πΆ as shown. We employ the following known triangle relationships:

πΆπΉ[π΄π΅πΆ]=π βπ=(ππ/2)sinπΎ=ππ

Just as [π΄π΅πΆ]=(ππ/2)sinπΎ, we have

πΆπΉπΆπΉπΆπΉπΆπΉ=[(πΆπΉ)(πΌπΆ)/2]sin(πΎ/2)=[π§cos(πΎ/2)][π§/2sin(πΎ/2)]=(π§2/2)cos(πΎ/2)sin(πΎ/2)=(π§2/4)sinπΎ

where we have used sin2πΎ= 2sinπΎcosπΎ in the last step.

Since πΆπΉπΌ is right, we have [πΆπΉπΌ]=π(π βπ)/2. Hence, we have two expressions for [π΄π΅πΆ]β2[πΆπΉπΌ]:

(ππ/2)sinπΎβ2(π§2/4)sinπΎ[(ππβπ§2/2]sinπΎ=ππ β2[π(π βπ/2]=ππ

as desired. Undefined control sequence \blacksquare

We now complete our proof. Dividing the result of Lemma 4 by (sinπΎ)/2 gives

ππβπ§2=2ππ/(sinπΎ).

Since Lemma 3 gives us ππβπ§2=π₯π¦π§/π and the Extended Law of Sines gives us sinπΎ=π/2π , the equation above becomes

π₯π¦π§/ππ₯π¦π§/ππ₯π¦π§=2ππ/(π/2π )=4π π=4π π2

Thus, we have shown that if πΌ is the incenter of triangle π΄π΅πΆ, we have

(πΌπ΄)(πΌπ΅)(πΌπΆ)=4π π2,

where π is the circumradius of π΄π΅πΆ and π is the inradius of π΄π΅πΆ.

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Note, we proved some intermediate results we probably didnβt have to (such as the fact that πΈ is on ray πΆπΌ) when the results were quick and easy to prove. Others we stated by fiat, such as [π΄π΅πΆ]=ππ , since the proofs are more involved, and we feel pretty safe that these results can be cited as known results without proof.

The above is a pretty daunting proof. What our solution doesnβt give is any indication of how we might have come up with this solution. If you didnβt find the above solution on your own, see if you can figure out how you might have come up with it now that you have seen it.

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