How to Write a Solution

Have a Plan
by Richard Rusczyk

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:

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

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 - 2r²/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*([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 - 2r²/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

Volume of ABCD = ([A] + [B] + [C] + [D])r/3 = rS/3

as desired.

---------------end lemma---------------

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 = (ha - 2r)/ha, or a = r - 2r2/ha.

(1)

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*([A]+[B]+[C]+[D])/S = 2r,

as desired.

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