Difference between revisions of "1969 IMO Problems/Problem 4"
Line 1: | Line 1: | ||
+ | ==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. | ||
+ | |||
+ | ==Solution== | ||
+ | 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>. | ||
+ | |||
+ | 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 | ||
+ | |||
+ | <math>r = \frac{|\triangle ABC|}{s} = \frac{ab}{a + b + c} = \frac{a + b - c}{2} = s - c</math> | ||
+ | |||
+ | 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 | ||
+ | |||
+ | <math>(a + b + c) (a + b - c) = a^2 + 2ab + b^2 - c^2 = 2ab</math> | ||
+ | |||
+ | or just because the angle <math>\angle C = 90^\circ</math> is right. Therefore, ST = 2r. Let R' be the midpoint of ST. Then | ||
+ | |||
+ | <math>AR' = AT - \frac{ST}{2} = b - r = b - \frac{a + b - c}{2} = \frac{c + b - a}{2} = s - a = AR</math> | ||
+ | |||
+ | 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: | ||
+ | |||
+ | <math>r_2 = SD = BS - BD = a - \frac{a^2}{c}</math> | ||
+ | |||
+ | <math>r_3 = TD = AT - AD = b - \frac{b^2}{c}</math> | ||
+ | |||
+ | 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 | ||
+ | |||
+ | <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> | ||
+ | |||
+ | 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. | ||
+ | |||
+ | The above solution was posted and copyrighted by yetti. The original thread can be found here: [https://aops.com/community/p376814] | ||
+ | |||
+ | == See Also == {{IMO box|year=1969|num-b=3|num-a=5}} |
Latest revision as of 13:42, 29 January 2021
Problem
A semicircular arc is drawn with
as diameter.
is a point on
other than
and
, and
is the foot of the perpendicular from
to
. We consider three circles,
, all tangent to the line
. Of these,
is inscribed in
, while
and
are both tangent to
and
, one on each side of
. Prove that
, and
have a second tangent in common.
Solution
Denote the triangle sides . Let
be the circumcircle of the right angle triangle
centered at the midpoint
of its hypotenuse
. Let
be the tangency points of the circles
with the line AB. In an inversion with the center
and positive power
(
being the inversion circle radius), the line AB is carried into itself, the circle
is carried into the altitude line
and the altitude line
into the circle
. This implies that the circle
intersecting the inversion circle
is carried into itself, but this is possible only if the circle
is perpendicular to the inversion circle
. It follows that the tangency point
of the circle
is the intersection of the inversion circle
with the line
. Similarly, in an inversion with the center B and positive power
(
being the inversion circle radius), the line AB is carried into itself, the circle
is carried into the altitude line
and the altitude line
into the circle
. This implies that the circle
intersecting the inversion circle
is carried into itself, but this is possible only if the circle
is perpendicular to the inversion circle
. It follows that the tangency point S of the circle
is the intersection of the inversion circle
with the line
.
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 of the right angle triangle
is equal to
where and s are the area and semiperimeter of the triangle
, for example, because of an obvious identity
or just because the angle is right. Therefore, ST = 2r. Let R' be the midpoint of ST. Then
Therefore, the points are identical and the midpoint of the segment ST is the tangency point R of the incircle
with the triangle side c = AB. It follows that the normals to the hypotenuse AB at the tangency points S, T of the circles
are tangent to the incircle
. Radii
of the circles
are now easily calculated:
Denote the centers of the circles
. The line
cuts the midline RI of the trapezoid
at the distance from the point R equal to
As a result, the centers are collinear (in fact, I is the midpoint of the segment
). The common center line
and the common external tangent AB of the circles
meet at their common external homothety center
and the other common external tangent of the circles
from the common homothety center H is a tangent to the circle
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 |