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Difference between revisions of "2021 JMC 10 Problems/Problem 21"

(Created page with "==Problem== Two identical circles <math>\omega_{a}</math> and <math>\omega_{b}</math> with radius <math>1</math> have centers that are <math>\tfrac{4}{3}</math> units apart. T...")
 
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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 $\omega_{a}$ and $\omega_{b}$ with radius $1$ have centers that are $\tfrac{4}{3}$ units apart. Two externally tangent circles $\omega_1$ and $\omega_2$ of radius $r_1$ and $r_2$ respectively are each internally tangent to both $\omega_a$ and $\omega_b$. If $r_1 + r_2 = \tfrac{1}{2}$, what is $r_1r_2$?

$\textbf{(A) }\dfrac{1}{21}\qquad\textbf{(B) }\dfrac{1}{14}\qquad\textbf{(C) }\dfrac{5}{63}\qquad\textbf{(D) }\dfrac{2}{21}\qquad\textbf{(E) }\dfrac{1}{7}$

Solution

Let $A$ and $B$ be the centers of $\omega_a$ and $\omega_b$ respectively. Let $R=1$ be the radius of $\omega_a$ and $\omega_b$ and $D=\tfrac{4}{3}$ be the distance between $A$ and $B$. Note that the centers of $\omega_1$ and $\omega_2$, say $O_1$ and $O_2$ respectively, lie on a line that is both perpendicular to $AB$ and equidistant from $A$ and $B$.


Because $\triangle AO_1 O_2 \cong \triangle BO_1 O_2$, we have that $\tfrac{D}{2}$ is the length of the $A$-altitude of $AO_1 O_2$. We have $AO_1 = R-r_1$, $AO_2 = R-r_2$, and $O_1 O_2 =r_1 +r_2$, so $\triangle AO_1 O_2$'s perimeter is $2R$. Thus, by Heron's Formula $[\triangle{AO_1 O_2}]=\sqrt{Rr_1 r_2 (R-r_1 -r_2)} =\tfrac{1}{2} \cdot \tfrac{D}{2} \cdot (r_1 + r_2)$. Substituting known values, we have \[\sqrt{1 \cdot r_1 r_2 (1-\tfrac{1}{2})} =\frac{\frac{4}{3}}{4} \cdot \frac{1}{2},\] whence $r_1 \cdot r_2= \tfrac{1}{18}$.

[asy] size(5cm); draw(circle((0,0),1/3),purple); draw(circle((0,-1/2),1/6),purple); draw(circle((-2/3,0),1)); draw(circle((2/3,0),1)); dot((0,0)); dot((0,-1/2)); dot((-2/3,0)); dot((2/3,0)); label("$A$", (-2/3,0), SW); label("$B$", (2/3,0), SE); draw((-2/3,0)--(2/3,0)); draw((-2/3,0)--(0,-1/2)--(2/3,0)); draw((0,-1/2)--(0,0), dotted); [/asy]

Remark: In this specific case, $\triangle{AO_1 O_2}$ is actually a right triangle with lengths in the ratio $3:4:5$, which is why the diagram has one of the centers lying on $AB$.

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