## FoolForMath 3 years ago Fool's problem of the day, (1) If $$p_1, p_2, p_3$$ are respectively the perpendicular from the vertices of $$\triangle ABC$$ to the opposite sides, then prove that $\hspace{3cm} \frac {\cos A}{p_1} + \frac {\cos B}{p_2} + \frac {\cos C}{p_3} = \frac 1 R$ where $$R$$ is the circumradius of the $$\triangle ABC$$ . Good luck!

1. Ishaan94

I am unable to draw me a nice diagram :(

2. FoolForMath

Diagram is not necessary to solve this problem.

3. blockcolder

4. FoolForMath

5. Henryblah

I think I saw something really similar to this on an AEA paper the other day...

6. Ishaan94

Why can't I solve it? What's making this question so hard to solve?

7. estudier

So far,I got 1/R = (CosA+CosB +CosC-1)/(p_1^-1+p_2^-1+p_3^-1)

8. FoolForMath

This requires some knowledge of certain other identities.

9. Ishaan94

Isn't Law of Sines enough? :D

10. estudier

No

11. estudier

You need things with R (and maybe inradius as well)

12. FoolForMath

I used $\sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C$

13. FoolForMath

and $\large R= \frac {abc}{4 \times Area}$

14. estudier

Yes, I was wondering about that last one.

15. estudier

I was trying to use like bc = 2R p1

16. Ishaan94

Using the Law of Sines, and Trig. substitution.$\frac3R = \frac{2\sin c \sin b }{p_1} + \frac{2\sin a \sin c}{p_2}+ \frac{2\sin b \sin a}{p_3}\tag 1$ $\frac3R = \underbrace{\frac{\cos a}{p_1}+\frac{\cos b}{p_2}+\frac{\cos c}{p_3}}_{x} + \frac{\cos (b-c)}{p_1}+\frac{\cos (a-c)}{p_2}+\frac{\cos (b-a)}{p_3}\tag 2$ $\small\frac3R = x + \underbrace{\frac{\sin c \sin b }{p_1} + \frac{\sin a \sin c}{p_2}+ \frac{\sin b \sin a}{p_3}}_{\frac{3}{R2},\quad \mathsf{\text{From (1)}}} + \frac{\cos c\cos b}{p_1}+\frac{\cos c\cos a}{p_2}+\frac{\cos b\cos a}{p_3}$ $\frac{3}{R} = x + \frac{\cos c\cos b}{p_1}+\frac{\cos c\cos a}{p_2}+\frac{\cos b\cos a}{p_3}\tag 3$ Something is wrong with my method, if you will compare (1),(2) and (3). You will see it. $2R = \frac{a}{\sin a}= \frac{b}{\sin b} = \frac{c}{\sin c} \mathsf{\tag {Law of Sines}}$ |dw:1335990248416:dw| $a=\frac{p_3}{\sin b}, \quad b =\frac{p_1}{\sin c},\quad c = \frac{p_2}{\sin a}.$

17. Ishaan94

:facepalm: Equation (3) is supposed to be, $$\frac{3}{2R} = x + \frac{\cos c\cos b}{p_1}+\frac{\cos c\cos a}{p_2}+\frac{\cos b\cos a}{p_3}\tag 3$$

18. Ishaan94

Ohh :( I thought maybe somehow trigonometric conversion will give me an answer, but I was so wrong.

19. Ishaan94

Using Co-ordinate Geometry, I somehow got an equation but it seems impossible to solve. $R^2 = \frac{\left( p_1 \sin 2a + p_3 \frac{\sin2c}{\sin b}\right)^2 + \left(2p_2 \cos b \cos a\right)^2}{(4\sin a \sin b \sin c)^2}$|dw:1335991322454:dw|Point B is at Origin.

20. Henryblah

All I get is... $cosA/P1 + cosB/P2 + cosC/P3=(P1^2+P2^2+P3^2)/(2P1P2P3)$ Then to prove the part after the equals is 1/R...

21. lgbasallote

has anyone solved it yet? :)

22. lgbasallote

awesome latex btw @Ishaan94

23. Ishaan94

I am missing on something, this isn't supposed to be this hard. :/ all the equations i am getting are way complicated. there must be a slick way to do it.

24. lgbasallote

yeah..you missed something alright ^_~

25. Ishaan94

lol

26. lgbasallote

i think you did something careless in your calculation :P

27. Ishaan94

From HenryBlah's equation.$\frac{p_1^2 + p_2^2 + p_3^2}{2p_1 p_2 p_3} = x$According to my notebook,$R = \frac{p_1p_2p_3}{4\sin a\sin b\sin c \Delta} \implies \frac{1}{p_1p_2p_3} = \frac{R}{4\sin a\sin b\sin c \Delta}$$\frac{p_1^2+p_2^2+p_3^2}{4\sin a\sin b\sin c \Delta} = \frac xR$

28. Ishaan94

I forgot the two lol

29. lgbasallote

haha see? :p

30. FoolForMath

You know I would have eliminated the $$p$$'s in the first step itself ;)

31. Ishaan94

|dw:1336017694001:dw|

32. Ishaan94

I can still eliminate them but I am trying to think what I am gonna do after it.

33. Ishaan94

$\frac{abc}{a\sin a \Delta} + \frac{abc}{\Delta c\sin c} + \frac{acb}{\Delta b\sin b} = 2\frac xR$$\Delta = \frac{abc}{4R}$ $\frac{1}{a\sin a} + \frac1{c\sin c} + \frac1{b\sin b} = \frac x {2R^2}$Where did I go wrong? I am not supposed to get this.

34. lgbasallote

can i show you my solution in messages fowwy? :)

35. Ishaan94

yeah sure

36. lgbasallote

lol :))