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## How to Write a Math Solution

##### January 1, 2012 by Richard Rusczyk and Matthew Crawford

You’ve figured out the solution to the problem—fantastic! But you’re 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 can’t 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 ‘How Not To’ solution and ‘How To’ solution. One common theme you’ll 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 ‘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 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.

### Have a Plan

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 you’ll 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 don’t skip anything and that you put your steps in an order that’s easy to follow.

Here’s 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:

$$a + b + c + d = 2r$$

Here’s a solution that looks short but is pretty tough to read:

How Not to Write the Solution:

Let our tetrahedron be $$ABCD$$. The small tetrahedron which includes vertex A is similar to the big tetrahedron. Since the face of this tetrahedron parallel to face $$BCD$$ is tangent to the sphere inscribed in $$ABCD$$, the distance between $$BCD$$ and this parallel face of the small tetrahedron is $$2r$$. Let’s call that small tetrahedron $$AXYZ$$. Hence, the altitude from $$A$$ in $$AXYZ$$ is $$h_a – 2r$$, where $$h_a$$ is the length of the altitude 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 similar with ratio $$a/r$$ (since that’s the ratio of the corresponding lengths, namely the radii of the inscribed spheres) we have $$a/r =$$ $$(h_a- 2r)/h_a$$.

The volume of the tetrahedron is $$[A]h_a/3$$, where $$[A]$$ is the area of triangle $$BCD$$. The volume of the tetrahedron can also be written $$rS/3$$, where $$S$$ is the surface area of $$ABCD$$. We can prove 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 find the volumes of the other 4 pieces. When we add them all up, we get
$\text{Volume of } ABCD = ([A] + [B] + [C] + [D]) r/3 = rS/3.$

We set that equal to our other volume expression and get $$h_a= rS/[A]$$. 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 get:

$$a = r – 2r[A]/S.$$

If we define $$[B]$$, $$[C]$$, and $$[D]$$ just like we defined $$[A]$$, we can use the same argument to get:

$$b = r – 2r[B]/S,$$$$c = r – 2r[C]/S,$$$$d = r – 2r[D]/S.$$

Adding these and our expression for $$A$$, we get:

$$a + b + c + d = 4r – 2r\cdot([A]+[B]+[C]+[D])/S = 2r,$$

as desired.

(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 $$rS/3$$. 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 won’t 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: $$ABCD, h_a, S, [A], AXYZ$$.

Order of things to prove:

1. Volume $$ABCD = rS/3$$ (lemma)
2. Show altitude $$AXYZ = h_a – 2r$$
3. Use similarity to get $$a = r – 2r^2/h_a$$
4. Equate volumes to get $$1/h_a= A/(rS)$$
5. sub $$4$$ into $$3$$ and add

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 ‘How Not to Write the Solution’.

How to Write the Solution:

Let our original tetrahedron be $$ABCD$$. We define:

$$[A]$$ = the area of the face of $$ABCD$$ opposite $$A$$
$$h_a$$ = the length of the altitude from $$A$$ to $$BCD$$
$$S$$ = the surface area of $$ABCD$$
$$AXYZ$$ = one of the small tetrahedrons formed as described

Define $$[B]$$, $$[C]$$, $$[D]$$ and $$h_b$$, $$h_c$$, $$h_d$$ similarly.

Lemma: The volume of tetrahedron $$ABCD$$ is given by $$rS/3$$.
Proof: Let $$I$$ be the center 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 can similarly find the volumes of the other 4 pieces. Adding these four tetrahedra gives us:
$\text{Volume of } ABCD = ABCD = ([A] + [B] + [C] + [D])r/3 = rS/3$
as desired.

Since face $$XYZ$$ of small tetrahedron $$AXYZ$$ is parallel to face $$BCD$$, tetrahedron $$AXYZ$$ is similar 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 distance 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 $$a = r – 2r^2/h_a$$.

The volume of the tetrahedron is $$h_a[A]/3$$. Setting this equal to the expression from Lemma 1 yields:

$$h_a = rS/[A]$$,

and substituting this into equation (1), we get:

$$a = r – 2r[A]/S$$.

By the same argument, we have:

$$b = r – 2r[B]/S,$$$$c = r – 2r[C]/S,$$$$d = r – 2r[D]/S.$$

Adding these and our expression for $$A$$, we get:

$$a + b + c + d = 4r – 2r\cdot ([A]+[B]+[C]+[D])/S = 2r,$$

as desired.

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

The first thing a reader sees on your paper isn’t the structure of your solution. It isn’t the answer, it isn’t the words you choose. It’s how the solution sits on the paper. If the reader has to decipher scrawl, you’re going to lose him. Ideally, you’ll typeset your solution with a program like LaTeX. However, in most contests you don’t have the luxury of turning to a computer and you’ll 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.

1. Use blank paper. Don’t use graph paper or lined paper – the lines often make solutions harder to read. Never use paper that is torn out of a spiral notebook.
2. 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.
3. 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.
4. Leave space at the top for a ‘Page _ of _’ so the reader knows how many pages there are, and what page she’s on. You probably won’t know how many pages you’ll write when you start, but you can fill these out when you’re finished. If you get to the bottom of a page and your solution must continue on another page, write ‘Continued’ at the bottom of that page so the reader knows we’re not 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, do not erase – the smudges from erasers make a mess.
7. When you make a mistake you’d like to omit, draw a single line through it and move on. If it’s a large block to omit, draw an ‘X’ through it and move on. Don’t scribble out large blocks of text.
8. 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 ‘Proof below.’ Below, you can write ‘(*) Addendum:’ and proceed with the proof. Don’t use a bunch of arrows to direct the reader all over the page.

Problem: Let $$S(n)$$ be the sum of the digits of $$n$$. Find

$$S(S(S(4444^{4444}))).$$

Below are two solutions. Neither solution is picture-perfect; when you’re under time pressure, it’s hard to write perfect-looking proofs. You should find the second one much more enjoyable to read. When you’re writing solutions, keep the above tips in mind, and just remember, ‘If they can’t read it, it’s not right.’

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:  ### U s e S p a c e

realizehowterriblyharditistotypelikethisitshardbecausewhenyoureusedtowritingcle
arlyandusingspaceandpunctuationandsentencestructureitgetsreallyhardtowritewith
outitsimilarlyonceyougetusedtoproperlyusingspaceinwritingyoursolutionsitwillbesec
ondnatureandyoullactuallyfindithardtowriteanindecipherableproof

When you write your solution you should:

1. Give each important definition or equation its own line.
2. Don’t bury too much algebra in a paragraph. You can write line after line of algebra, but put each step on its own line. Don’t cram the algebra in a paragraph.
3. Label equations or formulas or lemmas or cases you will use later very clearly.
4. Remember that there’s always more paper.

Problem: Let $$p(x)$$ be a polynomial with degree $$98$$ such that $$p(n) = 1/n$$ for $$n =$$ $$1, 2, 3,$$ $$4, \ldots , 99$$. Determine $$p(100)$$.

How Not to Write the Solution:

Let $$r(x) =$$ $$x (p(x) – 1/x)$$ $$= x p(x) – 1.$$ Since $$p(x)$$ is a polynomial with degree $$98$$, $$r(x)$$ is a polynomial with degree $$99$$. Since $$r(x) =$$ $$x (p(x) – 1/x),$$ and we are given that $$(p(x) – 1/x) = 0$$ for $$x = 1,$$ $$2, 3, \ldots , 99$$, $$r(x)$$ has roots $$1, 2, \ldots, 99.$$ Since $$r(x)$$ has degree $$99$$, these are the only roots of $$r(x)$$, which must thus have the form $$r(x) =$$ $$c(x – 1)(x – 2)$$ $$(x – 3) \cdots (x – 99)$$ for some constant $$c$$. To find $$c,$$ we first let $$x = 0$$ in equation $$r(x) =$$ $$x p(x) – 1,$$ yielding $$r(0) = -1.$$ Letting $$x = 0$$ in $$r(x) =$$ $$c(x – 1)(x – 2)$$ $$(x – 3) \cdots (x – 99)$$ yields $$r(0) = -c(99!);$$ hence, $$c = 1/99!.$$ Thus, we have $$r(x) =$$ $$(x – 1)(x – 2)$$ $$(x – 3) \cdots (x – 99)/99!.$$ We can combine the equations $$r(x) =$$ $$x p(x) – 1$$ and $$r(x) =$$ $$(x – 1)(x – 2)$$ $$(x – 3) \cdots (x – 99)/99!$$ and let $$x = 100$$ to find $$100p(100) – 1 =$$ $$(100 – 1)(100 – 2)$$ $$(100 – 3) \ldots(100 – 99)/99!,$$ so $$100p(100) – 1 =$$ $$99!/99! = 1,$$ so $$p(100) = 1/50.$$

Here’s the same solution, with nearly the same wording.

How to Write the Solution:

Let

$r(x)=x(p(x)-1/x)=xp(x)-1.\tag{1}$

Since $$p(x)$$ is a polynomial with degree $$98$$, $$r(x)$$ is a polynomial with degree $$99$$. Since $$r(x) =$$ $$x (p(x) – 1/x)$$, and we are given that $$(p(x) – 1/x) = 0$$ for $$x =$$ $$1, 2,$$ $$3, \ldots, 99,$$

$$r(x)$$ has roots $$1, 2, \ldots , 99.$$

Since $$r(x)$$ has degree $$99$$, these are the only roots of $$r(x)$$, which must thus have:

$r(x)=c(x-1)(x-2)(x-3)\cdots(x-99) \tag{2}$

for some constant $$c$$. To find $$c$$, we first let $$x = 0$$ in equation $$(1)$$, yielding $$r(0) = -1.$$ Letting $$x = 0$$ in $$(2)$$ yields $$r(0) = -c(99!)$$; hence, $$c = 1/99!$$. Thus, we have:

$$\begin{equation}r(x)=(x-1)(x-2)(x-3)\cdots(x-99)/99 ! \tag{3}\end{equation}$$

We can combine equations $$(1)$$ and $$(3)$$ and let $$x = 100$$ 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.$

### sdrawkcaB knihT, Write Forwards

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

“Figuring 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 eggs…”

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.

“Prepare 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.”

The reader doesn’t 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.

Here’s a sample problem:

Problem: Let $$a$$, $$b$$, 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 good way to see how we might come up with a solution from scratch, but it’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 it might be possible to manipulate 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. We begin by moving $$3abc$$ to the left hand side:

$$a^2(-a + b + c) = b^2(a – b + c) + c^2 (a + b – c) – 3abc \le 0.$$

If we 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 as to how the inequality can be reorganized. Since the inequality is cyclic, it seems natural to take these products in a way that preserves the cyclic nature. For instance, we 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 + ac^2 \ge 0.
\end{align*}

We see 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{align*}
(a – b)^2(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)^2(a – b + c) &= c^2(a – b + c) – 2ca(a – b + c) + a^2(a – b + c) \ge 0.\end{align*}

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^2(-a + b + c + a – b + c) \\
&- 2a^2b – 2ab^2 + 2abc + 2abc – 2b^2c – 2bc^2 – 2a^2c + 2abc – 2ac^2  \\
&= a^2(2a – 2b – 2c) + b^2(-2a + 2b – 2c) + c^2(-2a – 2b + 2c) + 6abc \ge 0.
\end{align*}

Multiplying this inequality by $$-1/2$$ reverses the inequality sign and gives us

$$a^2(-a + b + c) + b^2(a – b + c) + c^2(a + b – c) – 3abc \ge 0.$$

Adding $$3abc$$ to both sides gives us

$$a^2(-a + b + c) + b^2(a – b + c) + c^2(a + b – c) \ge 3abc$$

and we are done.

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

How to Write the Solution:

According to the triangle inequality, the sum of any two sides of a triangle is at least as great as the length of the third side. Thus, we have the inequalities

\begin{align*}
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,
\end{align*}

or

\begin{align*}
(b^2 – 2bc + c^2)(a – b – c)&\leq0, \\
(c^2- 2ac + a^2)(b – c – a) &\leq 0, \\
(a^2 – 2ab + b^2)(c – a – b)&\leq0.
\end{align*}

\begin{align*}
&a^2(a + b – c + a – b + c) + b^2(a + b – c – a + b + c) + c^2(-a + b + c + a – b + c) \\
&- 2a^2b – 2ab^2 + 2abc + 2abc – 2b^2c – 2bc^2 – 2a^2c + 2abc – 2ac^2  \\
&= a^2(2a – 2b – 2c) + b^2(-2a + 2b – 2c) + c^2(-2a – 2b + 2c) + 6abc \ge 0.
\end{align*}

Adding $$6abc$$ to both sides and dividing by $$2$$ we have the desired

$$a^2(-a + b + c) + b^2(a – b + c) + c^2(a + b – c)\leq3abc.$$

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 couldn’t 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 ‘A large thin-shelled vehicle for a young fowl that was created by a huge female bird,’ we call that big egg ‘Humpty-Dumpty’ 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 it’s 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 $$n$$ person is and what that $$f(x)$$ function stands for.

Here’s an example problem:

Problem: Show that for any set 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 because 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 order. If we start from the beginning, there are $$100$$ sums we can make by just 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 $$0$$, there are at most $$99$$ different remainders among these $$100$$ remainders. Therefore, two of these remainders must be the same since if there weren’t at least two the same there could only be $$99$$ total remainders (since we know none is zero). Now, take the difference between the two sums which have the same remainder when divided by $$100$$. This difference must have a remainder of $$0$$ when divided by $$100$$. Suppose we have subtracted the sum with fewer numbers from the one with more numbers. When we take this difference, all the numbers in the second sum cancel with numbers in the first sum, because each sum is just adding up numbers in our set starting with the first one but the second sum is shorter. Due to this cancellation, the difference of these two sums which have the same 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 by $$100$$, 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, 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:

Call the 100 integers $$n_1,$$ $$n_2,$$ $$\ldots,$$ $$n_{100}$$.

Let $$S_k = n_1 + n_2 + \dots + n_k$$ for $$k = 1,$$ $$2, \dots, 100.$$

Case 1: If $$S_1, S_2,\dots, S_{100}$$ are all distinct (mod 100), then exactly one of them must be a multiple of 100.

Case 2: Otherwise, the 100 sums, $$S_k$$, have at most 99 distinct residues (mod 100) and by the Pigeonhole Principle two of the sums, $$S_k$$, have the same residue (mod 100).

Thus means there exist 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 $$n_{j+1},$$ $$n_{j+2},$$ $$\ldots,$$ $$n_k$$. The sum of the elements of this subset is

\begin{align*}
& n_{j+1} + 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 Thousand Words

When you’re writing a solution to a geometry problem, or any problem involving a picture, you should include the diagram. If you don’t 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.

Here’s an example:

Problem: A point $$D$$ is placed on side $$BC$$ of triangle $$ABC$$. Circles are inscribed in $$ABD$$ and $$ACD$$. Their common exterior tangent (other than $$BC)$$ meets $$AD$$ at $$K$$. Prove that the length of $$AK$$ is independent of $$D$$.

Here’s a solution without a diagram:

How to Write the Solution:

Let our circles be $$O$$ and $$O’$$. Let $$M$$ and $$M’$$ be on $$O$$ and $$O’$$ such that $$MM’$$ is the common tangent through $$K$$. Let $$L$$ and $$L’$$ be the points where $$AD$$ meets circles $$O$$ and $$O’$$, respectively. Let $$N$$ 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 independent of $$D$$.

Since tangents from a point to a circle are equal, we 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 $$MM’ = NN’$$ by symmetry, we conclude that $$KL = DL$$. Hence,

$$MM’ = NN’ = DL + LL’ + KL’ = KD.$$

We 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 + AD + CD)}/2 – AC
\end{align*}

Thus,

$$NN’ = ND + DN’ = AD + {BC}/2 – {AC}/2 – {AB}/2.$$

Since $$NN’ = KD$$, we have

$$AD – KD = AK = {(AC + AB – BC)}/2,$$

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 member 3cnfsat in the Olympiad Geometry class)

Here’s a solution that includes the diagram:

How to Write the Solution: Let our circles be $$O$$ and $$O’$$. Let $$M$$ and $$M’$$ be on $$O$$ and $$O’$$ such that $$MM’$$ is the common tangent through $$K$$. Let $$L$$ and $$L’$$ be the points where $$AD$$ meets circles $$O$$ and $$O’$$, respectively. Let $$N$$ 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 independent of $$D$$.

Since tangents from a point to a circle are equal, we 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 $$MM’ = NN’$$ by symmetry, we conclude that $$KL = DL$$. Hence,

$$MM’ = NN’ = DL + LL’ + KL’ = KD.$$

We 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 + AD + CD)}/2 – AC
\end{align*}

Thus,

$$NN’ = ND + DN’ = AD + {BC}/2 – {AC}/2 – {AB}/2.$$

Since $$NN’ = KD$$, we have

$$AD – KD = AK = {(AC + AB – BC)}/2,$$

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 member 3cnfsat in the Olympiad Geometry class)

In our solution above, we used the fact that the length of the segments from a vertex of a triangle (like 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 $$DL$$ above) 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:

$$DN = (AB + AD + BD)/2 – AB$$

If you aren’t 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 ‘Why is that true?’ while reading your solution. She should also never be left in doubt as to whether or not you know why it is true.

It’s not always clear what steps you can assume the reader understands and what steps you have to explain. Here area few guidelines:

1. If you can cite a theorem that has a name, then you don’t have to prove the theorem. You can cite the theorem and move on, as 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 can also leave out the ‘By a well-known theorem’ bit, particularly for extremely common results such as the one just stated. (If you don’t know that result, try proving it on your own.)
3. If you still aren’t sure whether to prove a certain step or assume it’s well-known, you have a decision to make. If you can prove it in one or two lines, go ahead and do so. If it’s going to take a lot of work to prove but you know how to do it, then at least outline the proof (and give a more thorough one if you have time). If you’re taking a contest and have no idea how to prove it, cite it and move on. Maybe you’ll get lucky and it will be a ‘well-known theorem’. If you’re writing an educational paper and you don’t know how to prove it, then your paper isn’t finished until you figure it out.
4. When writing a string of algebraic steps, each step should follow obviously from the one before it. Don’t write something like, ‘Thus, we have $x^2(x – 4) + {(x +1)}^2 + 5x – 4(x +2) = -2,$ so $$x=1$$ is the only solution.’ You should include clear simple steps that make it clear that the above is equivalent to $$(x – 1)^3 = 0$$.
5. You can invoke symmetry or analogy when the cases are precisely the same. For example, suppose you want to prove that the area of any triangle $$ABC$$ is given by $[ABC] = (ab \sin C + bc \sin A + ac \sin B)/6,$ where $$a = BC$$, $$b = AC$$, and $$c = AB$$, and $$[ABC]$$ is the area of $$ABC$$. If you prove that $[ABC] = (ab/2)(\sin C),$ then you can just write, ‘Similarly, we have $[ABC] = (bc/2)(\sin A) = (ac/2)(\sin B).’$
6. When in doubt, explain it. Many of the solutions presented in this article have a little overkill in them. It’s better to prove too much than too little.

Here is a sample problem:

Problem: Find all integral solutions to the equation $$(m^2 + 1)(n^2 + 1) + 2(m – n)(1 – mn) = 4(mn + 1).$$ (Problem by Titu Andreescu).

This is totally unacceptable:

How Not to Write the Solution 1:

$$(1, 2), (-3, 0), (0, 3), (-2, -1), (1, 0), (-3, 2), (0, -1), (-2, 3)$$

The above is an answer, not a solution. This ‘solution’ lacks any evidence that these solutions actually work, and doesn’t 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 given equation rearranges to $$(m + 1)(n – 1) = \pm 2,$$ so the solutions are $(1, 2), (-3, 0), (0, 3), (-2, -1), (1, 0), (-3, 2), (0, -1), (-2, 3).$

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

How to Write the Solution:

We expand the first term and the right-hand side then regroup terms: $$(m^2+ 1)(n^2 + 1) + 2(m – n)(1 – mn) = 4(mn + 1)$$ $$m^2n^2+ m^2 + n^2+ 1 + 2(m – n)(1 – mn) = 4mn + 4$$ $$m^2n^2- 2mn + 1 + m^2 – 2mn + n^2 + 2(m – n)(1 – mn) = 4$$ $$(mn – 1)^2 + (m – n)^2 – 2(m – n)(mn – 1) = 4$$ This left side is the square of $$[(mn – 1) – (m – n)]$$, so we have: $$[(mn – 1)(m – n)]^2 = 4$$ $$[mn – m + n – 1]^2 = 4$$ $$[(m + 1)(n – 1)]^2 = 4$$ $$(m + 1)(n – 1) = \pm2,$$ Case 1: For $$(m + 1)(n – 1) = 2$$, we have the systems of equations: \begin{align*} m + 1 &= 1 \\ n – 1 &= 2 \\ &(0, 3) \end{align*} \begin{align*} m + 1 &= 2 \\ n – 1 &= 1 \\ &(1, 2) \end{align*} \begin{align*} m + 1 &= -1 \\ n – 1 &= -2 \\ &(-2, -1) \end{align*} \begin{align*} m + 1 &= -2 \\ n – 1 &= -1 \\ &(-3, 0) \end{align*} Case 2: For $$(m + 1)(n – 1) = -2$$, we have the systems of equations: \begin{align*} m + 1 &= 1 \\ n – 1 &= -2 \\ &(0, -1) \end{align*} \begin{align*} m + 1 &= -2 \\ n – 1 &= 1 \\ &(-3, 2) \end{align*} \begin{align*} m + 1 &= -1 \\ n – 1 &= 2 \\ &(-2, 3) \end{align*} \begin{align*} m + 1 &= 2 \\ n – 1 &= -1 \\ &(1, 0) \end{align*} Thus, the solutions are $$(1, 2),$$ $$(-3, 0),$$ $$(0, 3),$$ $$(-2, -1),$$ $$(1, 0),$$ $$(-3, 2),$$ $$(0, -1),$$ $$(-2, 3).$$

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 ‘Lemma’ and clearly delimiting the lemma and its proof from the rest of the solution.

Here’s a sample problem with two different solutions that employ lemmas. We’ve 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.

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

Here are two solutions:

How Not to Write the Solution 1: Let the line through $$A$$ parallel to $$BC$$ meet line $$BM$$ at $$J$$. Let the line through $$J$$ parallel to $$AB$$ meet line $$BC$$ at $$K$$. Let $$MN$$ hit $$AB$$ at $$X$$ and $$AC$$ at $$Y$$.

Since $$JK \parallel AB$$ and $$AJ \parallel BK$$, $$JKBA$$ is a parallelogram. Since

$$\angle{ABM} = (\angle A + \angle C)/2.$$

From right triangle $$BAM$$, we find

$$\angle BAM = 90 – \angle ABM = \angle B/2.$$

Since $$AJ \parallel BC$$, we have $$\angle JAB = \angle B$$, 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 rhombus.

The diagonals of a rhombus bisect each other, so triangles $$AMJ$$ and $$KMB$$ are congruent. Thus, the altitudes from $$M$$ to $$AJ$$ and $$BK$$ are equal and $$M$$ is equidistant from lines $$AJ$$ and line $$BK$$. By symmetry, this is also true for $$N$$. 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 have $$MX =$$ $$KB/2$$. Since $$JKBA$$ is a rhombus, $$KB = AB$$, so $$MX =$$ $$AB/2$$, as desired.

By symmetry, we also have $$NY =$$ $$AC/2$$. Since $$XY$$ is the midline of $$ABC$$ parallel to side $$BC$$, we have $$XY =$$ $$BC/2$$. Thus,

$$MN = MX + XY + YN = AB/2 + AC/2 + BC/2,$$

as desired.

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

### Clear Casework

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.

Here’s 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:

There 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 3 numbers like $$224$$. Also, there are $$111$$ and $$122$$ and $$133$$ and $$144$$ and so on. Each of those can be ordered in 3 ways, except for the $$111$$, which can be ordered in only one way. So, there are 52.

The solution above is short, and the answer is correct, but it’s not at all clear that all possibilities have been discovered. Also, it’s pretty tough to see that we have found exactly 52 solutions – 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:

We divide our investigation into cases based on the smallest digit of each number.

Case 1: 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 $$1 \leq n \leq 9$$ is any digit from $$1$$ to $$9$$. There are thus 9 numbers with smallest digit $$0$$ that satisfy the problem.

Case 2: The smallest digit is $$1$$.

If the smallest digit is $$1$$, the number must be of the form $$nn1$$, or permutations of this form (i.e. $$n1n$$ or $$1nn$$). However, these 3 permutations are the same when $$n = 1$$. Hence, we have 3 permutations each for $$2 \leq n \leq 9$$ and only $$1$$ for $$n = 1$$, for a total of $$1 +$$ $$3(8) =$$ $$\mathbf{25}$$ numbers with smallest digit $$1$$ that satisfy the problem.

Case 3: The smallest digit is $$2$$.

If the smallest digit is $$2$$, then the number is of the form $$2mn$$, where $$n = 2m$$, and permutations of this form. Our only options here are $$(m,n) = (2,4)$$, which gives us 3 numbers $$(224, 242, 422)$$, $$(m,n) = (3,6)$$, which gives us 6 numbers (permutations of $$236$$), and $$(m,n) = (4,8)$$, which also gives us 6 numbers. Hence, there are $$3 +$$ $$6 +$$ $$6 =$$ $$\mathbf{15}$$ numbers with smallest digit $$2$$.

Case 4: The smallest digit is $$3$$.

There are 3 solutions in this case: $$339$$, $$393$$, $$933$$.

Case 5: The smallest digit is larger than $$3$$.

If the smallest digit is larger than $$3$$, the smallest product we can form with two of the digits is $$4(4) = 16$$, which is not a single digit number. Hence, there are no numbers that satisfy the problem with smallest digit larger than $$3$$.

Since every possible 3-digit number falls in exactly one of these cases, we conclude that there are

$$9 + 25 + 15 + 3 = 52$$

numbers that satisfy the problem.

### Proofreed

Comunicacating complex 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 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 you’re 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 don’t want to write “abc + bcd” when you mean “abd + acd.” 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 “repetition is the mother of all skill.”

Problem: $$x$$, $$y$$, and $$z$$ are real numbers such that

$x + y + z = 5 \quad\text{and}\quad xy + yz + zx = 3.$

Determine with proof the largest value that any one of the three numbers can be.

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:

$(x + y)^2 = (5 – x)^2,$$xy = 3 – z(x + y) = 3 – z(5 – z).$

Now we note that

$$0 \le (x – y)^2 = (x + y) – 4xy.$$

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

$$0 \le (x + y)^2 – 4xy = 25 – 10z + z^2 – 12 + 20z – 4z^2 = 3z^2 + 10z + 13.$$

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

$$0 \ge -3z^2 + 10z + 13 = -(z + 1)(3z – 13).$$

The iequality holds when $$-1 \ge z \ge 13/3$$.

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

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:

$(x + y)^2 = (5 – x)^2,$$xy = 3 – z(x + y) = 3 – z(5 – z).$

Now we note that

$$0 \le (x – y)^2 = (x + y)^2 – 4xy.$$

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

$$0 \le (x + y)^2 – 4xy = 25 – 10z + z^2 – 12 + 20z – 4z^2= -3z^2 + 10z + 13.$$

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

$$0 ≤ -3z^2 + 10z + 13 = -(z + 1)(3z – 13).$$

The inequality holds when $$-1 ≤ z ≤ 13/3$$.

Since the given equations for $$x$$, $$y$$, and $$z$$ can be manipulated to form this same quadratic inequality in$$x$$, $$y$$, or $$z$$, they each have maximum possible values of $$13/3$$. This maximum can be achieved when $$x = y = 1/3$$ 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 fall over, then 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. Start off saying what you’re 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, ‘We will show by contradiction that there are infinitely many primes. Assume the opposite, that there are exactly $$n$$ primes $$\dots$$.’

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. ‘Thus, we have shown by contradiction that there are infinitely many prime numbers.’ You can also decorate the end of proofs with such items as $$QED$$ or $$AYD$$ or $$WWWWW$$ or $$\blacksquare$$ or //.

Here’s a sample problem:

Problem: Let $$I$$ be the incenter of triangle $$ABC$$. Prove

$$(IA) (IB) (IC) = 4r^2$$

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

(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 $$ABC$$.)

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

How Not to Write the Solution:

From $$\triangle AIC$$ we have $$\angle AIC =$$ $$180^{\circ} – \angle ACI – \angle CAI =$$ $$180^{\circ} – \alpha/2 – \gamma/2 =$$ $$180^{\circ} – (180^{\circ} – \beta)/2 =$$ $$90^{\circ} + \beta/2$$ and from $$\triangle EBC$$ we have $$\angle EBC =$$ $$\angle ABC + \angle ABQ =$$ $$\beta + (180^{\circ} – \beta)/2 =$$ $$90^{\circ} + \beta/2$$, so $$\angle AIC = \angle EBC$$. Thus, $$\triangle AIC \sim \triangle EBC$$ by Angle-Angle Similarity. By symmetry, we conclude $$\triangle BIC \sim \triangle EAC$$.

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

Draw altitude $$IE$$ of $$AIC$$. $$[EIC] =$$ $$(z^2/2) \sin \gamma =$$ $$r(s – c)/2$$, and $$[ABC] =$$ $$(ab/2) \sin \gamma =$$ $$rs$$, so $$[(ab – z^2)/2] \sin \gamma = rc$$. Then the Law of Sines and the earlier equation give our result.

Short, ugly, and completely incomprehensible.

How to Write the Solution: We let

$a = BC, b = AC, c = AB$$s = (a + b + c)/2$$\alpha = \angle BAC, \beta =\angle ABC, \gamma= \angle ABC,$
$[ABC] = \text{area of polygon }ABC$$x = IA, y = IB, z = IC$

Let the external bisectors of angles $$A$$ and $$B$$ of $$\triangle ABC$$ meet at $$E$$ as shown. Point $$E$$ is equidistant from lines $$AB$$, $$AC$$, and $$BC$$, so it is on angle bisector $$CI$$ as well. We will show

$[EACB] = bxy/2z + br/2 + a^2br/2z^2 = axy/2z + ar/2 + ab^2r/2z^2²,\tag{1}$

and

$(\sin \gamma)(ab – z^2)/2 = rc.\tag{2}$

From (1) we will show that $$ab – z^2 =$$ $$xyz / r$$, which we will combine with (2) and known triangle relationships to show the desired result.

Lemma 1: $$\triangle AIC \sim \triangle EBC$$ and $$\triangle BIC \sim \triangle EAC$$.
Proof: By symmetry, the two results are equivalent. We will show the first. Since $$CI$$ bisects $$\angle ACB$$, we have $$\angle ACI = \angle BCE$$.

From $$\triangle AIC$$ we have

$\angle AIC = 180^{\circ} – \angle ACI – \angle CAI = 180^{\circ} – \alpha/2 – \gamma/2 = 180^{\circ} – (180^{\circ} – \beta)/2 = 90^{\circ} + \beta/2$

and from $$\triangle EBC$$ we have

$\angle EBC = \angle ABC + \angle ABQ = \beta + (180^{\circ} – \beta)/2 = 90^{\circ} + \beta/2,$

so $$\triangle AIC \sim \triangle EBC$$ by Angle-Angle Similarity. By symmetry, we conclude $$\triangle AIC \sim \triangle EBC.$$ $$\blacksquare$$

Lemma 2: $[EACB] = bxy/2z + br/2 + a^2br/2z^2 = axy/2z + ar/2 + ab^2r/2z^2$ Proof: We find the area of $$EACB$$ by splitting it into pieces:

$$[EACB] = [AIE] + [AIC] + [EBC]$$

First we tackle $$[AIE]$$ by showing it is a right triangle with legs $$x$$ and $$by/z$$. From Lemma 1, we have $$\triangle BIC \sim \triangle EAC$$. Hence, $$AE/IB = AC/IC$$, or

$$AE = (AC)(IB)/IC = by/z.$$

Since

$[AIE] = (AI)(AE)/2 = (x)(by/z)/2 = bxy/2z.\tag{3}$

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

$[AIC] = br/2.\tag{4}$

Finally, since $$\triangle AIC \sim \triangle EBC$$, we have

$[EBC] = [AIC](BC/IC)^2 = (br/2)(a/z)^2 = a^2br/2z^2.\tag{5}$

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

$[EACB] = bxy/2z + br/2 + a^2br/2z^2\tag{6}$

By symmetry, we note that $$[EACB]$$ also equals our expression in (6) with $$a$$ and $$b$$ interchanged and $$x$$ and $$y$$ interchanged. Hence, we have the desired

$$[EACB] = bxy/2z + br/2 + a^2br/2z^2 = axy/2z + ar/2 + ab^2r/2z^2$$

$$\blacksquare$$

Lemma 3 : $$ab – z^2 = xyz / r$$.
Proof: Rearranging our result from Lemma 1 yields

$(bxy/2z + br/2 + a^2br/2z^2) – (axy/2z + ar/2 + ab^2r/2z^2) = 0$$(bxy/2z – axy/2z) + (br/2 – ar/2) + (a^2br/2z^2 – ab^2r/2z ^2) = 0$$(b-a)(xy/2z) + (b – a)(r/2) – (b – a) (abr/2z^2) = 0$$(b – a)(xy/2z + r/2 – abr/2z^2) = 0$

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

Case 1: $$b – a = 0$$. If $$b = a$$ then $$ABC$$ is isosceles and $$\alpha = \beta$$. Hence, the extension of angle bisector $$CI$$ is perpendicular to $$AB$$ at point $$D$$ as shown. Since $$I$$ is the incenter of $$ABC$$ and $$ID \perp AB$$, $$ID = r$$ since $$ID$$ is an inradius of $$ABC$$. Also,

$\angle IAB = \alpha/2 = \beta/2 = \angle IBA,$

so $$IB = IA$$ (i.e. $$x = y$$). Thus, the equation we wish to prove, $$ab – z^2 =$$ $$xyz / r$$, is in this case equivalent to

$a^2 – z^2 = x^2z/r. \tag{7}$

From right triangles $$CAD$$ and $$IAD$$, we have

$(c/2)^2 + r^2 = x^2, \tag{8}$$(c/2)^2 + (z + r)^2 = a^2. \tag{9}$

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

$c/2 = ar/z. \tag{10}$

Substituting (10) into (8) yields

$a^2r^2/z^2 + r^2 = x^2,$$a^2r^2 + r^2z^2 = x^2z^2,$$(r/z)(a^2 + z^2) = x^2z/r. \tag{11}$

Substituting (10) into (9) gives

$a^2r^2/z^2 + (z + r)^2 = a^2,$$a^2r^2 + z^2(z + r)^2 = a^2z^2,$$z^2(z + r)^2 = a^2z^2 – a^2r^2,$$z^2(z + r)^2 = a^2(z^2 – r^2),$$z^(z + r) = a^2(z – r),$$a^2r + z^2r = a^2z – z^3,$$(r/z)(a^2 + z^2) = a^2 – z^2. \tag{12}$

Combining (11) and (12) gives us the desired $$a^2 – z^2 = x^2z/r$$.

Case 2: $$xy / 2z + r / 2 – abr / 2z^2 = 0.$$

Multiplying this equation by $$2z^2 / r$$ yields

$$xyz / r + z^2 – ab = 0,$$

from which the desired $$ab – z^2 = xyz / r$$ immediately follows.

Thus, the lemma is proved. $$\blacksquare$$

Lemma 4: $$(\sin \gamma)(ab – z^2) / 2 = rc$$.

Proof: We draw altitude $$IF$$ perpendicular to $$BC$$ as shown. We employ the following known triangle relationships:

\begin{align*}
CF & = s – c \\
[ABC] & = (ab/2) \sin \gamma =rs
\end{align*}

Just as $$[ABC] = (ab/2) \sin \gamma$$, we have

\begin{align*}
CF & = [(CF)(IC)/2] \sin (\gamma/2) \\
CF & = [z \cos (\gamma /2)][z/2 \sin (\gamma/2)] \\
CF & = (z^2 / 2) \cos(\gamma / 2) \sin(\gamma / 2)\\
CF & = (z^2 /4) \sin \gamma
\end{align*}

where we have used $$\sin 2\gamma =$$ $$2 \sin \gamma \cos \gamma$$ in the last step.

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

\begin{align*}
(ab / 2) \sin \gamma – 2(z^2 / 4) \sin \gamma & = rs – 2[r(s – c / 2]\\
[(ab – z^2 / 2] \sin \gamma & = rs
\end{align*}

as desired. $$\blacksquare$$

We now complete our proof. Dividing the result of Lemma 4 by $$(\sin \gamma) / 2$$ gives

$$ab – z^2 = 2rc / (\sin \gamma).$$

Since Lemma 3 gives us $$ab – z^2 = xyz / r$$ and the Extended Law of Sines gives us $$\sin \gamma = c / 2R$$, the equation above becomes

\begin{align*}
xyz/r & = 2rc/(c/2R)\\
xyz/r & = 4Rr\\
xyz & = 4Rr^2
\end{align*}

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

$$(IA)(IB)(IC) = 4Rr^2,$$

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

Note, we proved some intermediate results we probably didn’t have to (such as the fact that $$E$$ is on ray $$CI$$) when the results were quick and easy to prove. Others we stated by fiat, such as $$[ABC]= rs$$, 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 doesn’t give is any indication of how we might have come up with this solution. If you didn’t 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.

Summary Article Name
How to Write a Math Solution
Description
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.
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AoPS News
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