- 06 Oct '07 17:12 / 1 editHere is a beautiful puzzle from geometry for all RHP members.

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. - 07 Oct '07 11:34B@St*** beat me!

180/7 is correct.

Call the angle in question 'a'

The final equation to solve is 7a/2 = 90 and in my drunkeness last night I made that a=45/7 instead of 180/7.

Proof

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 - 07 Oct '07 12:25
*Originally posted by wolfgang59***B@St*** beat me!**

Deceptively simple, isn't it? I started off looking for something much more complicated than that, using the sine rule etc. - 07 Oct '07 15:23

Right you are..nice and neat*Originally posted by wolfgang59***B@St*** beat me!**

180/7 is correct.

Call the angle in question 'a'

The final equation to solve is 7a/2 = 90 and in my drunkeness last night I made that a=45/7 instead of 180/7.

Proof

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 - ...[text shortened]... must equal the angle at Q

Theefore 90-3a/2 = 2a

90 = 7a/2

180 = 7a

180/7=a

QED