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How
to Write a Solution
Bookends
by
Richard Rusczyk
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. . . '
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 
or //.
Here's a sample problem:
(IA)(IB)(IC) = 4Rr2,
where R is the circumradius of ABC and r is the inradius of ABC.
(Problem due to 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.
AIC we have
<AIC =
180° - <ACI - <CAI = 180° - 
/2
- 
/2 = 180° -
(180° - 
)/2
= 90° + /2
and from EBC
we have <EBC
= <ABC + <ABQ =
+ (180° - )/2
= 90° + /2,
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/2)(a/z)2 = a2br/2z2, so [EACB] = bxy/2z + br/2 + a2br/2z2 = axy/2z + ar/2 + ab2r/2z2. Thus, (b - a)(xy/2z + r + abr/2z2) = 0 If b = a, then ab - z2 = xyz/r follows from the Pythagorean Theorem and the Angle Bisector Theorem. Otherwise, ab - z2 = xyz/r follows immediately.
Draw
altitude IE of AIC. [EIC] = (z2/2) sin
= r(s - c)/2, and [ABC] = (ab/2) sin
= rs, so [(ab - z2)/2] sin
= rc. Then the Law of Sines and the earlier equation give our result.
Short, ugly, and completely incomprehensible.
(click on image
to open the image in a separate window)
We let
a
= BC, b = AC, c = AB
s = (a + b + c)/2
= <BAC,
= <ABC, = <ACB
[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 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 + a2br/2z2 = axy/2z + ar/2
+ ab2r/2z2, |
(1) |
and
(sin
)(ab - z2)/2
= rc. |
(2) |
From (1) we will show that ab - z2 = xyz/r, which we will combine with (2) and known triangle relationships to show the desired result.
Lemma
1:
AIC ~ EBC
and BIC ~ EAC.
Proof: By symmetry, the two results are equivalent.
We will show the first. Since CI bisects <ACB, we have <ACI =
< BCE. From AIC
we have
<AIC
= 180° - <ACI - <CAI = 180° - /2
- /2 = 180° -
(180° - )/2
= 90° + /2
and
from EBC we have
<EBC
= <ABC + <ABQ =
+ (180° - )/2
= 90° + /2,
so
<AIC = <EBC. Thus, AIC
~ EBC by Angle-Angle
Similarity. By symmetry, we conclude BIC
~ EAC.
Lemma
2:
[EACB] = bxy/2z + br/2 + a2br/2z2 = axy/2z + ar/2
+ ab2r/2z2
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 BIC
~ EAC. Hence,
AE/IB = AC/IC, or
AE = (AC)(IB)/IC = by/z.
Since
<IAB = /2 and <BAE
= (180° - )/2
= 90° - /2, we
have <IAE = 90°. Hence,
[AIE]
= (AI)(AE)/2 = (x)(by/z)/2 = bxy/2z. |
(3) |
For triangle AIC we note that the altitude from I to AC is the inradius of ABC, so
[AIC]
= br/2. |
(4) |
Finally,
since AIC ~ EBC,
we have
[EBC]
= [AIC](BC/IC)2 = (br/2)(a/z)2 = a2br/2z2. |
(5) |
Adding (3), (4), and (5) yields
[EACB]
= bxy/2z + br/2 + a2br/2z2 |
(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 + a2br/2z2 = axy/2z + ar/2 + ab2r/2z2
Lemma
3 : ab - z2 = xyz/r.
Proof: Rearranging our result from Lemma 1 yields
(bxy/2z
+ br/2 + a2br/2z2) - (axy/2z + ar/2 + ab2r/2z2)
= 0
(bxy/2z - axy/2z)+ (br/2 - ar/2) + (a2br/2z2 -
ab2r/2z2) = 0
(b - a)(xy/2z) + (b - a)(r/2) -(b - a)(abr/2z2)
= 0
(b - a)(xy/2z + r/2 - abr/2z2) = 0
Thus, one of the terms in this product equals 0.
Case 1: b - a = 0.
(click on image
to open the image in a separate window)
If
b = a, then ABC is isosceles and
=.
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 
AB, ID = r since ID is an inradius of ABC. Also,
<IAB
= /2 =/2
= <IBA,
so IB = IA (i.e. x = y). Thus, the equation we wish to prove, ab - z2 = xyz/r, is in this case equivalent to
a2
- z2 = x2z/r |
(7) |
From right triangles CAD and IAD, we have
(c/2)2
+ r2 = x2 |
(8) |
| (c/2)2
+ (z + r)2 = a2 |
(9)
|
The Angle 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
+ r2 = x2 a2r2 + r2z2 = x2z2 (r/z)(a2 + z2) = x2z/r |
(11) |
Substituting (10) into (9) gives
a2r2/z2
+ (z + r)2 = a2 a2r2 + z2(z + r)2 = a2z2 z2(z + r)2 = a2z2 - a2r2 z2(z + r)2 = a2(z2 - r2) z2(z + r) = a2(z - r) a2r + z2r = a2z - z3 (r/z)(a2 + z2) = a2 - z2 |
(12) |
Combining (11) and (12) gives us the desired a2 - z2 = x2z/r.
Case 2:
xy/2z + 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.
Lemma
4: (sin )(ab
- z2)/2 = rc.
Proof:
(click
on image to open the image in a separate window)
We draw altitude IF perpendicular to BC as shown. We employ the following known triangle relationships:
CF = s - c
[ABC] = (ab/2) sin
= rs
Just
as [ABC] = (ab/2) sin ,
we have
[CFI]
= [(CF)(IC)/2] sin(/2)
= [z cos(/2)][z/2]
sin(/2)
=
(z2/2) cos(/2)
sin(/2)
= (z2/4) sin ,
where
we have used sin 2
= 2 sin cos
in the last step.
Since CFI is right, we have [CFI] = r(s - c)/2. Hence, we have two expressions for [ABC] - 2[CFI]:
(ab/2) sin -
2(z2/4)
sin = rs - 2[r(s
- c)/2]
[(ab - z2)/2] sin
= rc,
as
desired.
We
now complete our proof. Dividing the result of Lemma 4 by (sin )/2
gives
ab
- z2 = 2rc/(sin ).
Since
Lemma 3 gives us ab - z2 = xyz/r and the Extended Law of
Sines gives us sin
= c/2R, the equation above becomes
xyz/r
= 2rc/(c/2R)
xyz/r = 4Rr
xyz = 4Rr2
Thus, we have shown that if I is the incenter of triangle ABC, we have
(IA)(IB)(IC) = 4Rr2,
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.
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