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FoolForMath
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!
I am unable to draw me a nice diagram :(
Diagram is not necessary to solve this problem.
What's a circumradius?
@blockcolder: http://mathworld.wolfram.com/Circumradius.html
I think I saw something really similar to this on an AEA paper the other day...
Why can't I solve it? What's making this question so hard to solve?
So far,I got 1/R = (CosA+CosB +CosC-1)/(p_1^-1+p_2^-1+p_3^-1)
This requires some knowledge of certain other identities.
Isn't Law of Sines enough? :D
You need things with R (and maybe inradius as well)
I used \[\sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C \]
and \[ \large R= \frac {abc}{4 \times Area} \]
Yes, I was wondering about that last one.
I was trying to use like bc = 2R p1
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| \[a=\frac{p_3}{\sin b}, \quad b =\frac{p_1}{\sin c},\quad c = \frac{p_2}{\sin a}. \]
:facepalm: Equation (3) is supposed to be, \(\frac{3}{2R} = x + \frac{\cos c\cos b}{p_1}+\frac{\cos c\cos a}{p_2}+\frac{\cos b\cos a}{p_3}\tag 3\)
Ohh :( I thought maybe somehow trigonometric conversion will give me an answer, but I was so wrong.
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.
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...
has anyone solved it yet? :)
awesome latex btw @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.
yeah..you missed something alright ^_~
i think you did something careless in your calculation :P
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,\[R = \frac{p_1p_2p_3}{4\sin a\sin b\sin c \Delta} \implies \frac{1}{p_1p_2p_3} = \frac{R}{4\sin a\sin b\sin c \Delta}\]\[\frac{p_1^2+p_2^2+p_3^2}{4\sin a\sin b\sin c \Delta} = \frac xR\]
You know I would have eliminated the \(p\)'s in the first step itself ;)
I can still eliminate them but I am trying to think what I am gonna do after it.
\[\frac{abc}{a\sin a \Delta} + \frac{abc}{\Delta c\sin c} + \frac{acb}{\Delta b\sin b} = 2\frac xR\]\[\Delta = \frac{abc}{4R}\] \[ \frac{1}{a\sin a} + \frac1{c\sin c} + \frac1{b\sin b} = \frac x {2R^2}\]Where did I go wrong? I am not supposed to get this.
can i show you my solution in messages fowwy? :)