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Difference between revisions of "1969 IMO Problems/Problem 4"

(Created page with "A semicircular arc <math>\gamma</math> is drawn with <math>AB</math> as diameter. <math>C</math> is a point on <math>\gamma</math> other than <math>A</math> and <math>B</math>...")
 
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==Problem==
 
A semicircular arc <math>\gamma</math> is drawn with <math>AB</math> as diameter. <math>C</math> is a point on <math>\gamma</math> other than <math>A</math> and <math>B</math>, and <math>D</math> is the foot of the perpendicular from <math>C</math> to <math>AB</math>. We consider three circles, <math>\gamma_1, \gamma_2, \gamma_3</math>, all tangent to the line <math>AB</math>. Of these, <math>\gamma_1</math> is inscribed in <math>\triangle ABC</math>, while <math>\gamma_2</math> and <math>\gamma_3</math> are both tangent to <math>CD</math> and <math>\gamma</math>, one on each side of <math>CD</math>. Prove that <math>\gamma_1, \gamma_2</math>, and <math>\gamma_3</math> have a second tangent in common.
 
A semicircular arc <math>\gamma</math> is drawn with <math>AB</math> as diameter. <math>C</math> is a point on <math>\gamma</math> other than <math>A</math> and <math>B</math>, and <math>D</math> is the foot of the perpendicular from <math>C</math> to <math>AB</math>. We consider three circles, <math>\gamma_1, \gamma_2, \gamma_3</math>, all tangent to the line <math>AB</math>. Of these, <math>\gamma_1</math> is inscribed in <math>\triangle ABC</math>, while <math>\gamma_2</math> and <math>\gamma_3</math> are both tangent to <math>CD</math> and <math>\gamma</math>, one on each side of <math>CD</math>. Prove that <math>\gamma_1, \gamma_2</math>, and <math>\gamma_3</math> have a second tangent in common.
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==Solution==
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Denote the triangle sides <math>a = BC, b = CA, c = AB</math>. Let <math>\omega</math> be the circumcircle of the right angle triangle <math>\triangle ABC</math> centered at the midpoint <math>O</math> of its hypotenuse <math>c = AB</math>. Let <math>R, S, T</math> be the tangency points of the circles <math>K_1, K_2, K_3</math> with the line AB. In an inversion with the center <math>A</math> and positive power <math>r_A^2 = AC^2 = b^2</math> (<math>r_A</math> being the inversion circle radius), the line AB is carried into itself, the circle <math>\omega</math> is carried into the altitude line <math>CD</math> and the altitude line <math>CD</math> into the circle <math>\omega</math>. This implies that the circle <math>K_3</math> intersecting the inversion circle <math>A</math> is carried into itself, but this is possible only if the circle <math>K_3</math> is perpendicular to the inversion circle <math>A</math>. It follows that the tangency point <math>T</math> of the circle <math>K_3</math> is the intersection of the inversion circle <math>(A, r_A = b)</math> with the line <math>AB</math>. Similarly, in an inversion with the center B and positive power <math>r_B^2 = BC^2 = a^2</math> (<math>r_B</math> being the inversion circle radius), the line AB is carried into itself, the circle <math>\omega</math> is carried into the altitude line <math>CD</math> and the altitude line <math>CD</math> into the circle <math>\omega</math>. This implies that the circle <math>K_2</math> intersecting the inversion circle <math>B</math> is carried into itself, but this is possible only if the circle <math>K_2</math> is perpendicular to the inversion circle <math>B</math>. It follows that the tangency point S of the circle <math>K_2</math> is the intersection of the inversion circle <math>(B, r_B = a)</math> with the line <math>AB</math>.
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The distance between the tangency points S, T is the equal to ST = AT - AS = AT - (AB - BS) = AC - (AB - BC) = a + b - c. The radius r of the incircle <math>K_1</math> of the right angle triangle <math>\triangle ABC</math> is equal to
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<math>r = \frac{|\triangle ABC|}{s} = \frac{ab}{a + b + c} = \frac{a + b - c}{2} = s - c</math>
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where <math>|\triangle ABC|</math> and s are the area and semiperimeter of the triangle <math>\triangle ABC</math>, for example, because of an obvious identity
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<math>(a + b + c) (a + b - c) = a^2 + 2ab + b^2 - c^2 = 2ab</math>
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or just because the angle <math>\angle C = 90^\circ</math> is right. Therefore, ST = 2r. Let R' be the midpoint of ST. Then
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<math>AR' = AT - \frac{ST}{2} = b - r = b - \frac{a + b - c}{2} = \frac{c + b - a}{2} = s - a = AR</math>
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Therefore, the points <math>R' \equiv R</math> are identical and the midpoint of the segment ST is the tangency point R of the incircle <math>K_1</math> with the triangle side c = AB. It follows that the normals to the hypotenuse AB at the tangency points S, T of the circles <math>K_2, K_3</math> are tangent to the incircle <math>K_1</math>. Radii <math>r_2, r_3</math> of the circles <math>K_2, K_3</math> are now easily calculated:
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<math>r_2 = SD = BS - BD = a - \frac{a^2}{c}</math>
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<math>r_3 = TD = AT - AD = b - \frac{b^2}{c}</math>
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Denote <math>I, I_2, I_3</math> the centers of the circles <math>K_1, K_2, K_3</math>. The line <math>I_2I_3</math> cuts the midline RI of the trapezoid <math>STI_3I_2</math> at the distance from the point R equal to
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<math>\frac{SI_2 + TI_3}{2} = \frac{r_2 + r_3}{2} = \frac{a + b}{2} - \frac{a^2 + b^2}{2c} = \frac{a + b - c}{2} = r = RI</math>
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As a result, the centers <math>I_2, I, I_3</math> are collinear (in fact, I is the midpoint of the segment <math>I_2I_3</math>). The common center line <math>I_2I_3</math> and the common external tangent AB of the circles <math>K_1, K_2, K_3</math> meet at their common external homothety center <math>H \equiv I_2I_3 \cap AB</math> and the other common external tangent of the circles <math>K_2, K_3</math> from the common homothety center H is a tangent to the circle <math>K_1</math> as well.
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The above solution was posted and copyrighted by yetti. The original thread can be found here: [https://aops.com/community/p376814]
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== See Also == {{IMO box|year=1969|num-b=3|num-a=5}}

Revision as of 12:42, 29 January 2021

Problem

A semicircular arc $\gamma$ is drawn with $AB$ as diameter. $C$ is a point on $\gamma$ other than $A$ and $B$, and $D$ is the foot of the perpendicular from $C$ to $AB$. We consider three circles, $\gamma_1, \gamma_2, \gamma_3$, all tangent to the line $AB$. Of these, $\gamma_1$ is inscribed in $\triangle ABC$, while $\gamma_2$ and $\gamma_3$ are both tangent to $CD$ and $\gamma$, one on each side of $CD$. Prove that $\gamma_1, \gamma_2$, and $\gamma_3$ have a second tangent in common.

Solution

Denote the triangle sides $a = BC, b = CA, c = AB$. Let $\omega$ be the circumcircle of the right angle triangle $\triangle ABC$ centered at the midpoint $O$ of its hypotenuse $c = AB$. Let $R, S, T$ be the tangency points of the circles $K_1, K_2, K_3$ with the line AB. In an inversion with the center $A$ and positive power $r_A^2 = AC^2 = b^2$ ($r_A$ being the inversion circle radius), the line AB is carried into itself, the circle $\omega$ is carried into the altitude line $CD$ and the altitude line $CD$ into the circle $\omega$. This implies that the circle $K_3$ intersecting the inversion circle $A$ is carried into itself, but this is possible only if the circle $K_3$ is perpendicular to the inversion circle $A$. It follows that the tangency point $T$ of the circle $K_3$ is the intersection of the inversion circle $(A, r_A = b)$ with the line $AB$. Similarly, in an inversion with the center B and positive power $r_B^2 = BC^2 = a^2$ ($r_B$ being the inversion circle radius), the line AB is carried into itself, the circle $\omega$ is carried into the altitude line $CD$ and the altitude line $CD$ into the circle $\omega$. This implies that the circle $K_2$ intersecting the inversion circle $B$ is carried into itself, but this is possible only if the circle $K_2$ is perpendicular to the inversion circle $B$. It follows that the tangency point S of the circle $K_2$ is the intersection of the inversion circle $(B, r_B = a)$ with the line $AB$.

The distance between the tangency points S, T is the equal to ST = AT - AS = AT - (AB - BS) = AC - (AB - BC) = a + b - c. The radius r of the incircle $K_1$ of the right angle triangle $\triangle ABC$ is equal to

$r = \frac{|\triangle ABC|}{s} = \frac{ab}{a + b + c} = \frac{a + b - c}{2} = s - c$

where $|\triangle ABC|$ and s are the area and semiperimeter of the triangle $\triangle ABC$, for example, because of an obvious identity

$(a + b + c) (a + b - c) = a^2 + 2ab + b^2 - c^2 = 2ab$

or just because the angle $\angle C = 90^\circ$ is right. Therefore, ST = 2r. Let R' be the midpoint of ST. Then

$AR' = AT - \frac{ST}{2} = b - r = b - \frac{a + b - c}{2} = \frac{c + b - a}{2} = s - a = AR$

Therefore, the points $R' \equiv R$ are identical and the midpoint of the segment ST is the tangency point R of the incircle $K_1$ with the triangle side c = AB. It follows that the normals to the hypotenuse AB at the tangency points S, T of the circles $K_2, K_3$ are tangent to the incircle $K_1$. Radii $r_2, r_3$ of the circles $K_2, K_3$ are now easily calculated:

$r_2 = SD = BS - BD = a - \frac{a^2}{c}$

$r_3 = TD = AT - AD = b - \frac{b^2}{c}$

Denote $I, I_2, I_3$ the centers of the circles $K_1, K_2, K_3$. The line $I_2I_3$ cuts the midline RI of the trapezoid $STI_3I_2$ at the distance from the point R equal to

$\frac{SI_2 + TI_3}{2} = \frac{r_2 + r_3}{2} = \frac{a + b}{2} - \frac{a^2 + b^2}{2c} = \frac{a + b - c}{2} = r = RI$

As a result, the centers $I_2, I, I_3$ are collinear (in fact, I is the midpoint of the segment $I_2I_3$). The common center line $I_2I_3$ and the common external tangent AB of the circles $K_1, K_2, K_3$ meet at their common external homothety center $H \equiv I_2I_3 \cap AB$ and the other common external tangent of the circles $K_2, K_3$ from the common homothety center H is a tangent to the circle $K_1$ as well.

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

See Also

1969 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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