INMO 2015

5) Let ABCD be a convex quadrilateral.Let diagonals AC and BD intersect at P. Let PE,PF,PG and PH are altitudes from P on the side AB,BC,CD and DA respectively. Show that ABCD has a incircle if and only if $\frac{1}{PE}+\frac{1}{PG}=\frac{1}{PF}+\frac{1}{PH}$ .

1) Let ABC be a right-angled triangle with $\angle{B}=90^{\circ}$ . Let BD is the altitude from B on AC. Let P,Q and be the incenters of triangles ABD,CBD and ABC respectively. Show that circumcenter of triangle PIQ lie on the hypotenuse AC.

Easy Math Editor

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`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`	$\sum_{i=1}^3$
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Comments

solution are given in official hbsce site

onkar tiwari - 5 years, 5 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}`	$\boxed{123}$