## asnaseer Group Title Geometry challenge. Question to follow in drawing. 2 years ago 2 years ago

1. asnaseer

|dw:1328465419159:dw|

2. asnaseer

there are 3 squares next to each other. prove C = A + B

3. Mr.Math

3 squares with the same side length I presume?

4. asnaseer

yes - 3 identical squares

5. asnaseer

BTW: This is a geometry challenge - so no trig allowed :)

6. FoolForMath

Okay I can solve it using height and distance, one interesting fact: $$\sin(C-B) = \sin (C) \sin (B)$$

7. FoolForMath

Snap! asnaseer :P

8. Mr.Math

Oh man, this sucks!

9. asnaseer

:D - think "outside" the box

10. FoolForMath

Comone why not use trig? :P

11. asnaseer

ok - lets see the trig solution. but there is a very elegant geometric solution if you can find it. :)

12. FoolForMath

Hmm Asnaseer rules :D

13. asnaseer

I'm off to have some food - will be back soon ...

14. Mr.Math

I can't seem to be able to find a geometrical proof. Here's a proof using trigonometry: We have $$\sin C=\frac{1}{\sqrt{2}}$$, $$\sin B=\frac{1}{\sqrt{5}}$$, $$\sin A=\frac{1}{\sqrt{10}}$$, and $$\cos A=\frac{3}{\sqrt{10}}$$, $$\cos B=\frac{2}{\sqrt{5}}.$$ It's easy to show that $$\sin C=\sin(A+B)$$ and hence $$C=A+B \text{ because } A,B, C \in (0, \frac{\pi}{2}).$$

15. asnaseer

I'll let this problem /simmer/ for a while before showing the geometric proof - unless of course someone actually proves it by then :)

16. moneybird

any hints?

17. asnaseer

I gave a hint above - think "outside" the box :)

18. asnaseer

I guess I can give another hint to you guys, here it is...

19. asnaseer

|dw:1328468434372:dw| essentially, you want to try and prove that D=B

20. Shayaan_Mustafa

It can be proof using Euclidean concepts of geometry i.e. using theorems, postulates, axiom etc... It is actually difficult. But intresting.

21. asnaseer

There is actually a much easier proof using only the properties of similar triangles.

22. Shayaan_Mustafa

Yes I know in terms of Euclidean Geometry. These properties of triangles are followed by, called, postulates.