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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!
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!

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Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.1I am unable to draw me a nice diagram :(

FoolForMath
 3 years ago
Best ResponseYou've already chosen the best response.3Diagram is not necessary to solve this problem.

blockcolder
 3 years ago
Best ResponseYou've already chosen the best response.0What's a circumradius?

FoolForMath
 3 years ago
Best ResponseYou've already chosen the best response.3@blockcolder: http://mathworld.wolfram.com/Circumradius.html

Henryblah
 3 years ago
Best ResponseYou've already chosen the best response.0I think I saw something really similar to this on an AEA paper the other day...

Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.1Why can't I solve it? What's making this question so hard to solve?

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

FoolForMath
 3 years ago
Best ResponseYou've already chosen the best response.3This requires some knowledge of certain other identities.

Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.1Isn't Law of Sines enough? :D

estudier
 3 years ago
Best ResponseYou've already chosen the best response.0You need things with R (and maybe inradius as well)

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

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

estudier
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, I was wondering about that last one.

estudier
 3 years ago
Best ResponseYou've already chosen the best response.0I was trying to use like bc = 2R p1

Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.1Using 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 \[a=\frac{p_3}{\sin b}, \quad b =\frac{p_1}{\sin c},\quad c = \frac{p_2}{\sin a}. \]

Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.1:facepalm: Equation (3) is supposed to be, \(\frac{3}{2R} = x + \frac{\cos c\cos b}{p_1}+\frac{\cos c\cos a}{p_2}+\frac{\cos b\cos a}{p_3}\tag 3\)

Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.1Ohh :( I thought maybe somehow trigonometric conversion will give me an answer, but I was so wrong.

Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.1Using 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.

Henryblah
 3 years ago
Best ResponseYou've already chosen the best response.0All 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...

lgbasallote
 3 years ago
Best ResponseYou've already chosen the best response.0has anyone solved it yet? :)

lgbasallote
 3 years ago
Best ResponseYou've already chosen the best response.0awesome latex btw @Ishaan94

Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.1I 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.

lgbasallote
 3 years ago
Best ResponseYou've already chosen the best response.0yeah..you missed something alright ^_~

lgbasallote
 3 years ago
Best ResponseYou've already chosen the best response.0i think you did something careless in your calculation :P

Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.1From HenryBlah's equation.\[\frac{p_1^2 + p_2^2 + p_3^2}{2p_1 p_2 p_3} = x\]According to my notebook,\[R = \frac{p_1p_2p_3}{4\sin a\sin b\sin c \Delta} \implies \frac{1}{p_1p_2p_3} = \frac{R}{4\sin a\sin b\sin c \Delta}\]\[\frac{p_1^2+p_2^2+p_3^2}{4\sin a\sin b\sin c \Delta} = \frac xR\]

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

Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.1I can still eliminate them but I am trying to think what I am gonna do after it.

Ishaan94
 3 years ago
Best 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.

lgbasallote
 3 years ago
Best ResponseYou've already chosen the best response.0can i show you my solution in messages fowwy? :)
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