Question

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!

Answer 1

I am unable to draw me a nice diagram :(

Answer 2

Diagram is not necessary to solve this problem.

Answer 3

What's a circumradius?

Answer 4

@blockcolder: http://mathworld.wolfram.com/Circumradius.html

Answer 5

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

Answer 6

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

Answer 7

So far,I got

1/R = (CosA+CosB +CosC-1)/(p_1^-1+p_2^-1+p_3^-1)

Answer 8

This requires some knowledge of certain other identities.

Answer 9

Isn't Law of Sines enough? :D

Answer 10

No

Answer 11

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

Answer 12

I used \[\sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C \]

Answer 13

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

Answer 14

Yes, I was wondering about that last one.

Answer 15

I was trying to use like bc = 2R p1

Answer 16

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.
\[2R = \frac{a}{\sin a}= \frac{b}{\sin b} = \frac{c}{\sin c} \mathsf{\tag {Law of Sines}}\]


\[a=\frac{p_3}{\sin b}, \quad b =\frac{p_1}{\sin c},\quad c = \frac{p_2}{\sin a}. \]