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Difference between revisions of "2008 IMO Problems/Problem 6"

(Hints for problem 6 IMO 2008)
 
 
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==Problem==
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Let <math>ABCD</math> be a convex quadrilateral with <math>BA \ne BC</math>. Denote the incircles of triangles <math>ABC</math> and <math>ADC</math> by <math>\omega _1</math> and <math>\omega _2</math> respectively. Suppose that there exists a circle <math>\omega</math> tangent to ray <math>BA</math> beyond <math>A</math> and to the ray <math>BC</math> beyond <math>C</math>, which is also tangent to the lines <math>AD</math> and <math>CD</math>. Prove that the common external tangents to <math>\omega _1</math> and <math>\omega _2</math> intersect on <math>\omega</math>
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==Solution==
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{{solution}}
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Here are some hints:
 
Here are some hints:
  
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Vo Duc Dien
 
Vo Duc Dien
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==See Also==
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{{IMO box|year=2008|num-b=5|after=Last Problem}}

Latest revision as of 00:15, 19 November 2023

Problem

Let $ABCD$ be a convex quadrilateral with $BA \ne BC$. Denote the incircles of triangles $ABC$ and $ADC$ by $\omega _1$ and $\omega _2$ respectively. Suppose that there exists a circle $\omega$ tangent to ray $BA$ beyond $A$ and to the ray $BC$ beyond $C$, which is also tangent to the lines $AD$ and $CD$. Prove that the common external tangents to $\omega _1$ and $\omega _2$ intersect on $\omega$

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it.

Here are some hints:

Let B be the top vertex of triangle ABC, O and K are the centers of the incircles of triangles ABC and ADC with radii R and r, respectively. S is the center of the circumcircle tangential to the extensions of AB, AC and DC. And let

E be the foot of the projection of O to AB. U be the foot of the projection of O to AC. V be the foot of the projection of K to AC. M be the foot of the projection of S to DC. L be the foot of the projection of K to DC. L be the intercept of DC and AB.

We have

/_EOB = /_LSC /_AOU = /_SOC /_ASO = /_KSC /_ASK = /_OSC /_LSB = /_KCO UV = BC – AB AU = VC OA. AK. cos(/_OAK) = OC. KC. cos(/_OCK) OK**2 = (R + r)**2 + UV**2 SK**2 = (R' + r)**2 + ML**2

Use sin (90-x) = cos x and cos(90-x) = sinx and characteristic of triangle a**2 = b**2 + c**2 - 2.b.c.cosine(angle) to solve.

Vo Duc Dien

See Also

2008 IMO (Problems) • Resources
Preceded by
Problem 5 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Last Problem
All IMO Problems and Solutions
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