Difference between revisions of "2021 JMC 10 Problems/Problem 21"
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Let <math>A</math> and <math>B</math> be the centers of <math>\omega_a</math> and <math>\omega_b</math> respectively. Let <math>R=1</math> be the radius of <math>\omega_a</math> and <math>\omega_b</math> and <math>D=\tfrac{4}{3}</math> be the distance between <math>A</math> and <math>B</math>. Note that the centers of <math>\omega_1</math> and <math>\omega_2</math>, say <math>O_1</math> and <math>O_2</math> respectively, lie on a line that is both perpendicular to <math>AB</math> and equidistant from <math>A</math> and <math>B</math>. | Let <math>A</math> and <math>B</math> be the centers of <math>\omega_a</math> and <math>\omega_b</math> respectively. Let <math>R=1</math> be the radius of <math>\omega_a</math> and <math>\omega_b</math> and <math>D=\tfrac{4}{3}</math> be the distance between <math>A</math> and <math>B</math>. Note that the centers of <math>\omega_1</math> and <math>\omega_2</math>, say <math>O_1</math> and <math>O_2</math> respectively, lie on a line that is both perpendicular to <math>AB</math> and equidistant from <math>A</math> and <math>B</math>. | ||
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Because <math>\triangle AO_1 O_2 \cong \triangle BO_1 O_2</math>, we have that <math>\tfrac{D}{2}</math> is the length of the <math>A</math>-altitude of <math>AO_1 O_2</math>. We have <math>AO_1 = R-r_1</math>, <math>AO_2 = R-r_2</math>, and <math>O_1 O_2 =r_1 +r_2</math>, so <math>\triangle AO_1 O_2</math>'s perimeter is <math>2R</math>. Thus, by Heron's Formula <math>[\triangle{AO_1 O_2}]=\sqrt{Rr_1 r_2 (R-r_1 -r_2)} =\tfrac{1}{2} \cdot \tfrac{D}{2} \cdot (r_1 + r_2)</math>. Substituting known values, we have <cmath>\sqrt{1 \cdot r_1 r_2 (1-\tfrac{1}{2})} =\frac{\frac{4}{3}}{4} \cdot \frac{1}{2},</cmath> whence <math>r_1 \cdot r_2= \tfrac{1}{18}</math>. | Because <math>\triangle AO_1 O_2 \cong \triangle BO_1 O_2</math>, we have that <math>\tfrac{D}{2}</math> is the length of the <math>A</math>-altitude of <math>AO_1 O_2</math>. We have <math>AO_1 = R-r_1</math>, <math>AO_2 = R-r_2</math>, and <math>O_1 O_2 =r_1 +r_2</math>, so <math>\triangle AO_1 O_2</math>'s perimeter is <math>2R</math>. Thus, by Heron's Formula <math>[\triangle{AO_1 O_2}]=\sqrt{Rr_1 r_2 (R-r_1 -r_2)} =\tfrac{1}{2} \cdot \tfrac{D}{2} \cdot (r_1 + r_2)</math>. Substituting known values, we have <cmath>\sqrt{1 \cdot r_1 r_2 (1-\tfrac{1}{2})} =\frac{\frac{4}{3}}{4} \cdot \frac{1}{2},</cmath> whence <math>r_1 \cdot r_2= \tfrac{1}{18}</math>. |
Latest revision as of 15:56, 1 April 2021
Problem
Two identical circles and
with radius
have centers that are
units apart. Two externally tangent circles
and
of radius
and
respectively are each internally tangent to both
and
. If
, what is
?
Solution
Let and
be the centers of
and
respectively. Let
be the radius of
and
and
be the distance between
and
. Note that the centers of
and
, say
and
respectively, lie on a line that is both perpendicular to
and equidistant from
and
.
Because , we have that
is the length of the
-altitude of
. We have
,
, and
, so
's perimeter is
. Thus, by Heron's Formula
. Substituting known values, we have
whence
.
Remark: In this specific case, is actually a right triangle with lengths in the ratio
, which is why the diagram has one of the centers lying on
.