# Try to crack this ...isosceles triangle.

ranjan sinha
Posers and Puzzles 06 Oct '07 17:12
1. 06 Oct '07 17:121 edit
Here 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. ðŸ˜³ðŸ˜²
2. wolfgang59
06 Oct '07 17:29
I have succeeded.
3. 06 Oct '07 18:261 edit
Originally posted by wolfgang59
I have succeeded.
4. wolfgang59
06 Oct '07 19:27
No solution.

Just succeeded at trying to crack it. (As post challenged!) ðŸ˜‰

Seriously though I did try and as with most of these triangle puzzles I just went around in circles and proved the triangle had 180degrees!

Good Luck to anyone else who tries.
5. wolfgang59
06 Oct '07 20:54
45/7 degrees

Obvious ðŸ˜€
6. 07 Oct '07 06:58
Originally posted by wolfgang59
45/7 degrees

Obvious ðŸ˜€
Nope that's not correct..
7. wolfgang59
07 Oct '07 09:51
This has been driving me crazy! All I have succeeded in doing is
* proving that the angle equals itself
* proving that there are 180 degress in the triangle
* designing a new Star Trek logo

You will have to PM me answer if I dont get it in next 24 hours!
8. 07 Oct '07 11:081 edit
180/7.

I'll come back and give the reasoning later, if it's right.
9. wolfgang59
07 Oct '07 11:34
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 -[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
10. 07 Oct '07 12:25
Originally posted by wolfgang59
B@St*** beat me!
ðŸ™‚

Deceptively simple, isn't it? I started off looking for something much more complicated than that, using the sine rule etc.
11. wolfgang59
07 Oct '07 12:42
Interestingly this is the same triangle from another (much harder) post.

12. 07 Oct '07 15:20
Originally posted by wolfgang59
Interestingly this is the same triangle from another (much harder) post.

It's the same answer. Not exactly the same triangle.
Nice solution!
13. 07 Oct '07 15:23
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
Right you are..nice and neat
14. 07 Oct '07 15:281 edit
Originally posted by luskin
It's the same answer. Not exactly the same triangle.
Nice solution!
puzlles are different . Solutions are the same though...This puzzle is simpler than the one in the thread quoted.
15. wolfgang59