**Problem**: (1986 China TST - #5) Given a square $ABCD$ whose side length is $1$, $P$ and $Q$ are points on the sides $AB$ and $AD$, respectively. If the perimeter of $APQ$ is $2$ find the angle $PCQ$.

**Solution**: Let $ AP = x, AQ = y $ and $ \angle PCB = \alpha, \angle QCD = \beta $. By the given condition, we have $ AP+AQ+QP = x+y+\sqrt{x^2+y^2} = 2 $ (1). From this, we find

$ x+y = 2-\sqrt{x^2+y^2} $

$ (x+y)^2 = 4-4\sqrt{x^2+y^2}+(x^2+y^2) $

$ xy = 2-2\sqrt{x^2+y^2} $

Substituting $ \sqrt{x^2+y^2} = 2-x-y $ from (1), we have

$ xy = 2-2(2-x-y) \Rightarrow 2-x-y = x+y-xy \Rightarrow \frac{2-x-y}{x+y-xy} = 1 $ (2).

Note that $ \tan{\alpha} = \frac{PB}{BC} = 1-x $ and $ \tan{\beta} = \frac{QD}{DC} = 1-y $. Then

$ \tan{(\alpha+\beta)} = \frac{\tan{\alpha}+\tan{\beta}}{1-\tan{\alpha} \cdot \tan{\beta}} = \frac{(1-x)+(1-y)}{1-(1-x)(1-y)} = \frac{2-x-y}{x+y-xy} $.

But by (2), we have $ \tan{(\alpha+\beta)} = \frac{2-x-y}{x+y-xy} = 1 \Rightarrow \alpha+\beta = 45^{\circ} $.

Hence $ \angle PCQ = 90^{\circ}-(\alpha+\beta) = 45^{\circ} $. QED.

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Comment: Trigonometry can come in handy quite often, especially when dealing with angles. Purely geometric solutions to this problem are a lot more complicated in my opinion; when in doubt, use algebra. The following identity comes in handy on quite a few problems.

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Practice Problem: Prove that in any triangle we have $ \tan{\frac{A}{2}} \cdot \tan{\frac{B}{2}} + \tan{\frac{B}{2}} \cdot \tan{\frac{C}{2}} + \tan{\frac{C}{2}} \cdot \tan{\frac{A}{2}} = 1 $.

Inscribing a circle into triangle ABC, and using a = x + y, b = y + z, c = z + x, and r, we find that the identity is equivalent to

ReplyDeleter^2 / xz + r^2 / xy + r^2 / yz = 1

By Heron's, the area of the triangle A = \sqrt{xyz(x+y+z)} = (x+y+z)r, so A^2 / A^2 = 1 = r^2 (x + y + z) / xyz = 1, which is equivalent to the above identity. QED.

Don't you think you should start to slow down a little. USAMO is in two days!!!

ReplyDeleteTomorrow is my cool down period for the year :).

ReplyDeleteHaha, nice. Well, good luck on the USAMO. You are a great role model. You've worked hard and improved so much in one year.

ReplyDeleteI wonder if this proof is valid:

ReplyDeleteFrom the equations AP + AQ + PQ = 2, AP = 1 - PB, and AQ = 1 - QD we get

PQ = PB + QD

This means that triangles PCB and QCD can be "flipped" inward along PC and QC respectively to fill up triangle PCQ.

Then we have 2(alpha + beta) = 90, or alpha + beta = 45 which means alpha + beta = angle PCQ = 45.

That's interesting. Intuitively, it makes sense, but I'm not completely sure if I would be able to rigorously prove it. A good idea, though!

ReplyDeleteHmm, I guess "flipping triangles" inward is not the best way to put it.

ReplyDeleteHow about:

rotate triangle PCB so that BC lines up with DC. Then the newly formed triangle

"PCQDB" is SSS congruent with the original PCQ, so alpha + beta = angle PCQ, and alpha+beta + angle PCQ = 90, so PCQ = 45

I think you can prove it... draw an altitude down from c to pq and use algebra about the triangle areas and stuff

ReplyDelete