Orthocentre Problem

If $x, y, z$ are the distances of the vertices of the points $A,B$ and $C$ respectively in $ΔABC$ from the orthocentre, then simplify the expression:

$\frac{a}{x}+\frac{b}{y}+\frac{c}{z}$

Easy Math Editor

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Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

We know that : $x = 2RcosA$ , $y = 2RcosB$ and $z = 2RcosC$

$\frac{a}{x} + \frac{b}{y} + \frac{c}{z}$

= $\frac{2RsinA}{2RcosA} + \frac{2RsinB}{2RcosB} + \frac{2RsinC}{2RcosC}$

= $tanA + tanB + tanC$

jatin yadav - 7 years, 8 months ago

Nice attempt.. But if the options were the following then what would be the solution???:

$(a)\frac{xyz}{abc}$

$(b)\frac{(xy+yz+zx)^{3}}{(abc)^{2}}$

$(c)\frac{abc}{xyz}$

Krishna Jha - 7 years, 8 months ago

i think its option- c. since A+B+C=180. THUS tanA +tanB +tanC =tanAtanBtanC. but now clearly tanA=(a/x) tanB=(b/y) and tanC=(c/z). therefore, (a/x)+(b/y)+(c/z)=(abc/xyz). which is a pretty obvious.

Shubhabrota Chakraborty - 7 years, 8 months ago

Thanks.. but could you tell me a good book/website would i find these formulae(those of properties of triangles) with their proof??

Krishna Jha - 7 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$