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Difference between revisions of "1978 IMO Problems/Problem 4"

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==Problem==
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In a triangle <math>ABC</math> we have <math>AB = AC.</math> A circle which is internally tangent with the circumscribed circle of the triangle is also tangent to the sides <math>AB, AC</math> in the points <math>P,</math> respectively <math>Q.</math> Prove that the midpoint of <math>PQ</math> is the center of the inscribed circle of the triangle <math>ABC.</math>
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==Solution==
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Denote <math>a = BC, b = AB = AC</math> the triangle sides, <math>r, R</math> the triangle inradius and circumradius and <math>s</math>, <math>\triangle</math> the triangle semiperimeter and area. Let <math>S, T</math> be the tangency points of the incircle (I) with the sides <math>AB, AC</math>. Let <math>K, L</math> be the midpoints of of <math>BC, PQ</math> and <math>M</math> the midpoint of the arc <math>BC</math> of the circumcircle (O) opposite to the vertex <math>A</math>. The isosceles triangles <math>\triangle AST \sim \triangle APQ \sim \triangle ABC</math> are all centrally similar with the homothety center <math>a</math>. The homothety coefficient for the triangles <math>\triangle AST \sim \triangle ABC</math> is <math>h_{13} = \frac{AS}{AB} = \frac{s - a}{b}</math>. The homothety coefficient for the triangles <math>\triangle AST \sim \triangle APQ</math> is <math>h_{12} = \frac{AS}{AP}</math>. The point K is the tangency point of the incircle (I) with the side BC, while the point M is the tangency point of the given circle with the triangle circumcircle (O). It follows that the triangles <math>\triangle ASK \sim \triangle APM</math> are also centrally similar with the homothety center A and the same homothety coefficient as the triangles <math>\triangle AST \sim \triangle APQ</math>: <math>h_{12} = \frac{AS}{AP} = \frac{AK}{AM} = \frac{h}{2R}</math>, where h = AK is the A-altitude of the triangle <math>\triangle ABC</math>. The homothety coefficient of the triangles <math>\triangle APQ \sim \triangle ABC</math> is then <math>h_{23} = \frac{h_{12}}{h_{13}} = \frac{s - a}{b} \cdot \frac{2R}{h}</math> Denote <math>h' = AL</math> the <math>A</math>-altitude of the triangle <math>\triangle APQ</math>. Then
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<math>\frac{h'}{h} = h_{23} = \frac{s - a}{b} \cdot \frac{2R}{h}</math>
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<math>KL = h - h' = h - 2R\ \frac{s - a}{b}</math>
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Substituting <math>h = \frac{2 \triangle}{a}</math>, <math>2R = \frac{ab^2}{2 \triangle}</math> and <math>s - a = \frac{\triangle^2}{s(s - b)^2}</math>, we get
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<math>KL = \frac{2 \triangle}{a} - \frac{ab^2}{2 \triangle} \cdot \frac{\triangle^2}{bs(s - b)^2} =</math>
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(substituting 2s = 2b + a and <math>s - b = \frac a 2</math> for an isosceles triangle)
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<math>= \frac{\triangle}{s}\left(\frac{2s}{a} - \frac{ab}{2(s - b)^2}\right) = \frac{\triangle}{s} \left(\frac{2b}{a} + 1 - \frac{2b}{a}\right) = \frac{\triangle}{s} = r</math>
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which means that the point <math>L \equiv I</math> is identical with the incenter of the triangle <math>\triangle ABC</math>.
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The above solution was posted and copyrighted by yetti. The original thread for this problem can be found here: [https://aops.com/community/p367852]
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== See Also == {{IMO box|year=1978|num-b=3|num-a=5}}

Latest revision as of 17:04, 29 January 2021

Problem

In a triangle $ABC$ we have $AB = AC.$ A circle which is internally tangent with the circumscribed circle of the triangle is also tangent to the sides $AB, AC$ in the points $P,$ respectively $Q.$ Prove that the midpoint of $PQ$ is the center of the inscribed circle of the triangle $ABC.$

Solution

Denote $a = BC, b = AB = AC$ the triangle sides, $r, R$ the triangle inradius and circumradius and $s$, $\triangle$ the triangle semiperimeter and area. Let $S, T$ be the tangency points of the incircle (I) with the sides $AB, AC$. Let $K, L$ be the midpoints of of $BC, PQ$ and $M$ the midpoint of the arc $BC$ of the circumcircle (O) opposite to the vertex $A$. The isosceles triangles $\triangle AST \sim \triangle APQ \sim \triangle ABC$ are all centrally similar with the homothety center $a$. The homothety coefficient for the triangles $\triangle AST \sim \triangle ABC$ is $h_{13} = \frac{AS}{AB} = \frac{s - a}{b}$. The homothety coefficient for the triangles $\triangle AST \sim \triangle APQ$ is $h_{12} = \frac{AS}{AP}$. The point K is the tangency point of the incircle (I) with the side BC, while the point M is the tangency point of the given circle with the triangle circumcircle (O). It follows that the triangles $\triangle ASK \sim \triangle APM$ are also centrally similar with the homothety center A and the same homothety coefficient as the triangles $\triangle AST \sim \triangle APQ$: $h_{12} = \frac{AS}{AP} = \frac{AK}{AM} = \frac{h}{2R}$, where h = AK is the A-altitude of the triangle $\triangle ABC$. The homothety coefficient of the triangles $\triangle APQ \sim \triangle ABC$ is then $h_{23} = \frac{h_{12}}{h_{13}} = \frac{s - a}{b} \cdot \frac{2R}{h}$ Denote $h' = AL$ the $A$-altitude of the triangle $\triangle APQ$. Then

$\frac{h'}{h} = h_{23} = \frac{s - a}{b} \cdot \frac{2R}{h}$

$KL = h - h' = h - 2R\ \frac{s - a}{b}$

Substituting $h = \frac{2 \triangle}{a}$, $2R = \frac{ab^2}{2 \triangle}$ and $s - a = \frac{\triangle^2}{s(s - b)^2}$, we get

$KL = \frac{2 \triangle}{a} - \frac{ab^2}{2 \triangle} \cdot \frac{\triangle^2}{bs(s - b)^2} =$

(substituting 2s = 2b + a and $s - b = \frac a 2$ for an isosceles triangle)

$= \frac{\triangle}{s}\left(\frac{2s}{a} - \frac{ab}{2(s - b)^2}\right) = \frac{\triangle}{s} \left(\frac{2b}{a} + 1 - \frac{2b}{a}\right) = \frac{\triangle}{s} = r$

which means that the point $L \equiv I$ is identical with the incenter of the triangle $\triangle ABC$.

The above solution was posted and copyrighted by yetti. The original thread for this problem can be found here: [1]

See Also

1978 IMO (Problems) • Resources
Preceded by
Problem 3 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 5
All IMO Problems and Solutions
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