1. ## geometry olympics 2014

https://postimg.org/image/epnzx6clr/

I couldn't solve that and need help.

2. ## Re: geometry olympics 2014

I get $x = 6 \pm 2\sqrt{2}$

3. ## Re: geometry olympics 2014

Answer is 12 according to the book.How did you find it?
I triedsimilar triangles but no result.

4. ## Re: geometry olympics 2014

Can you show your attempt using similar triangles?

5. ## Re: geometry olympics 2014

Originally Posted by kastamonu
Answer is 12 according to the book.How did you find it?
I triedsimilar triangles but no result.
yes, the sum of x's is $(6+2\sqrt{2})+(6-2\sqrt{2}) = 12$

First, I drew a picture to draw in an auxiliary line segment & label some variables ...

equations I came up with using Pythagoras & similar triangles ...

$x^2+2^2=(b\sqrt{2})^2$

$a^2+b^2=12^2$

$x^2+14^2 = (a+b)^2$

$\dfrac{x}{a+b}=\dfrac{b}{12}$

Found an equation quadratic in $x$ and solved ...

6. ## Re: geometry olympics 2014

Please can you show the quadratic equation? I worked a lot on that one.

7. ## Re: geometry olympics 2014

Originally Posted by kastamonu
Please can you show the quadratic equation? I worked a lot on that one.
using the first Pythagoras equation, $b^2 = \dfrac{x^2+4}{2}$

using the second & third Pythagoras equations, $x^2+14^2 = 12^2 + 2ab \implies ab = \dfrac{x^2+52}{2}$

using the proportion from similar triangles, $12x = ab+b^2$

substitute for $ab$ and $b^2$ ... you get an equation quadratic in $x$

many thanks.

9. ## Re: geometry olympics 2014

I will disturb you one more time.
https://postimg.org/image/rzkw43x1l/

10. ## Re: geometry olympics 2014

Originally Posted by kastamonu
I will disturb you one more time.
https://postimg.org/image/rzkw43x1l/
Draw an auxiliary line between B and E ... pretty much the same equations as your last problem. You'll get a quadratic in $x$. Choose the larger of the two solutions. Do it.

11. ## Re: geometry olympics 2014

Many Thanks.x is 3.