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

1. asnaseer Group Title

|dw:1328465419159:dw|

2. asnaseer Group Title

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

3. Mr.Math Group Title

3 squares with the same side length I presume?

4. asnaseer Group Title

yes - 3 identical squares

5. asnaseer Group Title

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

6. FoolForMath Group Title

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

7. FoolForMath Group Title

Snap! asnaseer :P

8. Mr.Math Group Title

Oh man, this sucks!

9. asnaseer Group Title

:D - think "outside" the box

10. FoolForMath Group Title

Comone why not use trig? :P

11. asnaseer Group Title

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

12. FoolForMath Group Title

Hmm Asnaseer rules :D

13. asnaseer Group Title

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

14. Mr.Math Group Title

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 Group Title

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 Group Title

any hints?

17. asnaseer Group Title

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

18. asnaseer Group Title

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

19. asnaseer Group Title

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

20. Shayaan_Mustafa Group Title

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

21. asnaseer Group Title

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

22. Shayaan_Mustafa Group Title

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