ABC is an isosceles triangle with AB = AC. A perpendicular PQ is drawn on AB at its mid-point P (i.e. AP =PB), and this perpendicular line cuts AC at Q. It turns out that QC = BC. What is the the vertex angle /_BAC? The solution is to be obtained analytically and exactly. No trigonometrical tables..No numerical solutions. I assure you it is analytically solvable.

I spent literally 40 hours solving it, using all that I learned in high school starting from trig and ending in comparing areas btween the curves. Trust me, you really don't need this. It's solvable by pure geometry and algebra. Well, and some usage of your brain, of course:)) If you have an answer - write it here and PM me the solution, I will PM you if you were right, or respond in the topic, if you were not.

I have worked out an answer, and I've PM'd it to you. Took me about 20 minutes using my calculator and a bit of knowledge.

It's Wrong^)

Pfft Shows what use A-Level Maths is then! Interested to find out what the actual answer is then.

The isoceles triangle ABC has angle at B & C of (90-a/2)
Now consider the isoceles triangle QBC. It has angle at C of (90-a/2)
The angle at Q is {180 -[90-a] - [90-a]} = 2a
The angle at B is {[90-a/2]-a} = {90-3a/2}
But the angle at B must equal the angle at Q
Theefore 90-3a/2 = 2a
90 = 7a/2
180 = 7a
180/7=a

QED

That answer is the same answer I gave you



I think we might have worked through it slightly differently, but we do in fact get the same answer.