Difference between revisions of "1978 IMO Problems/Problem 4"
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− | + | ==Problem== | |
+ | 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> | ||
+ | |||
+ | ==Solution== | ||
+ | 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 | ||
+ | |||
+ | <math>\frac{h'}{h} = h_{23} = \frac{s - a}{b} \cdot \frac{2R}{h}</math> | ||
+ | |||
+ | <math>KL = h - h' = h - 2R\ \frac{s - a}{b}</math> | ||
+ | |||
+ | 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 | ||
+ | |||
+ | <math>KL = \frac{2 \triangle}{a} - \frac{ab^2}{2 \triangle} \cdot \frac{\triangle^2}{bs(s - b)^2} =</math> | ||
+ | |||
+ | (substituting 2s = 2b + a and <math>s - b = \frac a 2</math> for an isosceles triangle) | ||
+ | |||
+ | <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> | ||
+ | |||
+ | which means that the point <math>L \equiv I</math> is identical with the incenter of the triangle <math>\triangle ABC</math>. |
Revision as of 17:03, 29 January 2021
Problem
In a triangle we have
A circle which is internally tangent with the circumscribed circle of the triangle is also tangent to the sides
in the points
respectively
Prove that the midpoint of
is the center of the inscribed circle of the triangle
Solution
Denote the triangle sides,
the triangle inradius and circumradius and
,
the triangle semiperimeter and area. Let
be the tangency points of the incircle (I) with the sides
. Let
be the midpoints of of
and
the midpoint of the arc
of the circumcircle (O) opposite to the vertex
. The isosceles triangles
are all centrally similar with the homothety center
. The homothety coefficient for the triangles
is
. The homothety coefficient for the triangles
is
. 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
are also centrally similar with the homothety center A and the same homothety coefficient as the triangles
:
, where h = AK is the A-altitude of the triangle
. The homothety coefficient of the triangles
is then
Denote
the
-altitude of the triangle
. Then
Substituting ,
and
, we get
(substituting 2s = 2b + a and for an isosceles triangle)
which means that the point is identical with the incenter of the triangle
.