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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!
 one year ago
 one year 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!
 one year ago
 one year ago

This Question is Closed

Ishaan94Best ResponseYou've already chosen the best response.1
I am unable to draw me a nice diagram :(
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.3
Diagram is not necessary to solve this problem.
 one year ago

blockcolderBest ResponseYou've already chosen the best response.0
What's a circumradius?
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.3
@blockcolder: http://mathworld.wolfram.com/Circumradius.html
 one year ago

HenryblahBest ResponseYou've already chosen the best response.0
I think I saw something really similar to this on an AEA paper the other day...
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.1
Why can't I solve it? What's making this question so hard to solve?
 one year ago

estudierBest ResponseYou've already chosen the best response.0
So far,I got 1/R = (CosA+CosB +CosC1)/(p_1^1+p_2^1+p_3^1)
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.3
This requires some knowledge of certain other identities.
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.1
Isn't Law of Sines enough? :D
 one year ago

estudierBest ResponseYou've already chosen the best response.0
You need things with R (and maybe inradius as well)
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.3
I used \[\sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C \]
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.3
and \[ \large R= \frac {abc}{4 \times Area} \]
 one year ago

estudierBest ResponseYou've already chosen the best response.0
Yes, I was wondering about that last one.
 one year ago

estudierBest ResponseYou've already chosen the best response.0
I was trying to use like bc = 2R p1
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.1
Using the Law of Sines, and Trig. substitution.\[\frac3R = \frac{2\sin c \sin b }{p_1} + \frac{2\sin a \sin c}{p_2}+ \frac{2\sin b \sin a}{p_3}\tag 1\] \[\frac3R = \underbrace{\frac{\cos a}{p_1}+\frac{\cos b}{p_2}+\frac{\cos c}{p_3}}_{x} + \frac{\cos (bc)}{p_1}+\frac{\cos (ac)}{p_2}+\frac{\cos (ba)}{p_3}\tag 2\] \[\small\frac3R = x + \underbrace{\frac{\sin c \sin b }{p_1} + \frac{\sin a \sin c}{p_2}+ \frac{\sin b \sin a}{p_3}}_{\frac{3}{R2},\quad \mathsf{\text{From (1)}}} + \frac{\cos c\cos b}{p_1}+\frac{\cos c\cos a}{p_2}+\frac{\cos b\cos a}{p_3}\] \[\frac{3}{R} = x + \frac{\cos c\cos b}{p_1}+\frac{\cos c\cos a}{p_2}+\frac{\cos b\cos a}{p_3}\tag 3\] 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 \[a=\frac{p_3}{\sin b}, \quad b =\frac{p_1}{\sin c},\quad c = \frac{p_2}{\sin a}. \]
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.1
:facepalm: Equation (3) is supposed to be, \(\frac{3}{2R} = x + \frac{\cos c\cos b}{p_1}+\frac{\cos c\cos a}{p_2}+\frac{\cos b\cos a}{p_3}\tag 3\)
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.1
Ohh :( I thought maybe somehow trigonometric conversion will give me an answer, but I was so wrong.
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.1
Using Coordinate 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:dwPoint B is at Origin.
 one year ago

HenryblahBest ResponseYou've already chosen the best response.0
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...
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
has anyone solved it yet? :)
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
awesome latex btw @Ishaan94
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.1
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.
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
yeah..you missed something alright ^_~
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
i think you did something careless in your calculation :P
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.1
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,\[R = \frac{p_1p_2p_3}{4\sin a\sin b\sin c \Delta} \implies \frac{1}{p_1p_2p_3} = \frac{R}{4\sin a\sin b\sin c \Delta}\]\[\frac{p_1^2+p_2^2+p_3^2}{4\sin a\sin b\sin c \Delta} = \frac xR\]
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.3
You know I would have eliminated the \(p\)'s in the first step itself ;)
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.1
dw:1336017694001:dw
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.1
I can still eliminate them but I am trying to think what I am gonna do after it.
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.1
\[\frac{abc}{a\sin a \Delta} + \frac{abc}{\Delta c\sin c} + \frac{acb}{\Delta b\sin b} = 2\frac xR\]\[\Delta = \frac{abc}{4R}\] \[ \frac{1}{a\sin a} + \frac1{c\sin c} + \frac1{b\sin b} = \frac x {2R^2}\]Where did I go wrong? I am not supposed to get this.
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
can i show you my solution in messages fowwy? :)
 one year ago
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