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A candidate problem for 2008 Beijing middle school exam

Posted June 25th, 2008 by Mgccl

in

High school in China includes year 10 to 12. Students who want to get into a good high school have to take the middle school exam and score amazing on it. Each province or sometimes, cities, have their own exam.

This is one of the candidate problem for Beijing this year. It is suppose to be one of those questions that fail 99% of the student body in order to allow only elite to remain.
I saw it here[Chinese].
I tried the question and I was shocked how I suck at geometry. So I used a trigonometry solution, and in the end, using calculus.

Question: Find $\angle 7$ .

My solution:
$\angle 1 = \tan^{-1}(\frac{b}{2a+b})$
$\angle 2 = \tan^{-1}(\frac{a}{a+b})$
$\angle 3 = \tan^{-1}(\frac{2a+b}{a+b})$
$\angle 4 = \tan^{-1}(\frac{a+b}{2a+b})$
$\angle 5 = \angle 4 - \angle 1$
$\angle 6 = \angle 3 - \angle 2$
$\angle 7 = \angle 5 + \angle 6$
$\angle 7 = \tan^{-1}(\frac{a+b}{2a+b}) - \tan^{-1}(\frac{b}{2a+b}) + \tan^{-1}(\frac{2a+b}{a+b}) - \tan^{-1}(\frac{a}{a+b})$

Great, we have the expression! Solving it is the problem
$\angle 7 = \left(\tan^{-1}(\frac{a+b}{2a+b}) + \tan^{-1}(\frac{2a+b}{a+b})\right)- \left(\tan^{-1}(\frac{b}{2a+b}) \tan^{-1}(\frac{a}{a+b})\right)$
The formula
$\tan^{-1}(\frac{1}{x}) = \frac{\pi}{2} - \tan^{-1}(x)$
helps.
$\angle 7 = \frac{\pi}{2}-\left(\tan^{-1}(\frac{b}{2a+b}) + \tan^{-1}(\frac{a}{a+b})\right)$
How can I do the rest? There are two options, both run some huge calculations.
One is to use the tangent addition formula which I forgot and have to derive from sine and cosine addition formulas... but that will be too long for tests like this in China. I didn't Wikipedia it, it's cheating.
So I use calculus!
$f(b) = \tan^{-1}(\frac{b}{2a+b}) + \tan^{-1}(\frac{a}{a+b})$
$f(0) = \frac{\pi}{4}$
f'(x) = 0 (huge amount of calculation omitted, I believe it's around the same if I use the tangent addition formula if I remembered it.
$\angle 7 = \frac{\pi}{2} - \frac{\pi}{4}$
$\angle 7 = \frac{\pi}{4}$

There is a elegant geometry solution[Chinese]. This problem cost me few hours, most of the time spending on search for a geometric solution. I finally give up at around 3:30 AM and go with the brutal force "Find all angles and bash the numbers out of arctans."
I would totally fail the real test. I'm questioning myself do I have math talent at all...
Have to improve my geometry skills.

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Indeed, the solution based

On Jun/27/2008 03:36:19 Dave S (not verified) says:

Indeed, the solution based on geometry is much more elegant than the trig approach.

But considering the fact that geometry is notorious for its clumsiness in relating angles to lengths, Solving by trig is a natural and straightforward approach to the problem. If I was asked to solve the problem in a set time frame, say 10 minutes, I would choose the trig method as you did.

reply

hehe, to do these problems

On Jun/27/2008 12:13:34 Mgccl says:

hehe, to do these problems with all trig... then I have to be like god at finding trig functions include knowing all the identities.

reply

i did it the cheap way

On Sep/03/2008 19:03:07 Anonymous says:

i did it the cheap way haha
for this question to be valid, it must be true for all a/b's
therefore, set a to 0.00000000001, and b to 100000000000000000000
and u have a+b = 2a+b
EPC = BAC = 45 degrees