Affine transformations technique (Putnam 2001, A-4)

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I am trying to learn the technique of affine transformations from this article. The first question covered is question A4 on the Putnam of 2001.

(Putnam 2001, A4) $\triangle ABC$ has area one. Point $E$, $F$, $G$ lie on $BC$, $CA$, and $AB$ respectively such that $AE$ bisects $BF$ at point $R$, $BF$ bisects $CG$ at $S$, and $CG$ bisects $AE$ at $T$. Find the area of $\triangle RST$.

Problem configuration. Figure not drawn to scale.

By affine transformations we can take $\triangle ABC$ to be equilateral or right-isosceles as we see fit. When $\triangle ABC$ is right, we have that $\frac{AG}{AB} = \frac{BE}{BC} = \frac{CF}{CA} = r$. This makes sense. But then it gets crazy. Apparently, in the right-isosceles case, "we can use the fact that $CG$ bisects $AE$ to obtain the identity $(1 – r)(1 – \frac{r}{2}) = 1/2$". Why is this? (Later on there are other perplexing claims such as: $\frac{CT}{CG} = \frac{1}{2(1-r)}$ and $BS = SG$, but hopefully if I can understand how the author comes up with one of them then the others will become more apparent).

I looked up the official Putnam solution and they seemed to have use the affine technique slightly differently. Solution two (of six) uses affine transformation to take $\triangle ABC$ into a particular triangle with area one (namely the one with vertices $(0,1)$, $(1,0)$, and $(-1,0)$. By co-linearity of subsets of these points, we can further come up with three equations in three unknowns (these equations are not linear, but are still solvable). Plugging the values in for our three unknowns gives us the coordinates of the points $R$, $S$, and $T$. We can use the Shoelace Lemma to find the area of $\triangle RST$. Since the area of $\triangle ABC$ is already one, any affine transformation will leave the ratio of the areas of $\triangle ABC$ to $\triangle RST$ fixed. The only issue with the approach is the shear amount of computation required if one where to solve the problem by hand.

Best Answer

Here's how we can get $(1-r)(1 - \frac12 r) = \frac12$.

Drop the altitude from $T$ onto $BC$; let $H$ be the base of that altitude.

  • On the one hand, $\triangle THE$ is similar to $\triangle ABE$, and since $T$ bisects $AE$, we know that $\triangle THE = \frac12 \triangle ABE$. In particular, $TH = \frac12 AB$, and $HE = \frac12 BE = \frac r2 BC$.
  • On the other hand, $\triangle THC$ is similar to $\triangle GBC$. From above, we know that $HC = (1 - \frac r2)BC$, so $TH = (1 - \frac r2) GB = (1 - \frac r2)(1 - r)AB$.

This gives us two expressions for $TH$ in terms of $AB$, so we conclude that $(1 - \frac r2)(1-r) = \frac 12$.

(The "isosceles right triangle" thing is a matter of taste. It makes it easy to say "drop the altitude", but we could have gotten the same effect in any triangle by drawing a line through $T$ parallel to $AB$, and letting $H$ be its intersection with $BC$.)

The ratio $\frac{CT}{CG}$ also comes from the similarity of $\triangle THC$ and $\triangle GBC$, while to get $BS = SG$ (something that's only true in the right-isosceles case, since affine transformations don't preserve ratios of non-parallel segments!) we must drop the altitude from $S$ onto $AB$ and see that it bisects $BG$.

I'd add that once you know $r$ (which there are many ways to get), I see easier ways to finish the proof. Since $AT = TE$, we have $[ATC] = [TEC]$, where brackets denote area. In the equilateral case, we have $[AFST] = [CERS]$ by symmetry; by subtracting, we get $[CFS] = [RST]$.

So let $a = [AGT] = [BER] = [CFS] = [RST]$ and let $b = [AFST] = [BGTR] = [CERS]$. We are given in the problem that $4a+3b = [ABC] = 1$; meanwhile, $2a+b = [AGC] = r$. This gives us two equations to solve for $a$ and $b$, and $a$ is what we want to find.

(In general, I highly recommend tracking ratios of areas in affine transformation proofs.)

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