Difference between revisions of "2021 AIME I Problems/Problem 13"
Sugar rush (talk | contribs) |
MRENTHUSIASM (talk | contribs) m (→Solution) |
||
(15 intermediate revisions by 4 users not shown) | |||
Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
− | Circles <math>\omega_1</math> and <math>\omega_2</math> with radii <math>961</math> and <math>625</math>, respectively, intersect at distinct points <math>A</math> and <math>B</math>. A third circle <math>\omega</math> is externally tangent to both <math>\omega_1</math> and <math>\omega_2</math>. Suppose line <math>AB</math> intersects <math>\omega</math> at two points <math>P</math> and <math>Q</math> such that the measure of minor arc <math>\widehat{PQ}</math> is <math>120^{\circ}</math>. | + | Circles <math>\omega_1</math> and <math>\omega_2</math> with radii <math>961</math> and <math>625</math>, respectively, intersect at distinct points <math>A</math> and <math>B</math>. A third circle <math>\omega</math> is externally tangent to both <math>\omega_1</math> and <math>\omega_2</math>. Suppose line <math>AB</math> intersects <math>\omega</math> at two points <math>P</math> and <math>Q</math> such that the measure of minor arc <math>\widehat{PQ}</math> is <math>120^{\circ}</math>. Find the distance between the centers of <math>\omega_1</math> and <math>\omega_2</math>. |
− | ==Solution | + | ==Solution 1== |
− | Let <math>O_i</math> and <math>r_i</math> be the center and radius <math>\omega_i</math>, and let <math>O</math> and <math>r</math> be the center and radius of <math>\omega</math>. | + | Let <math>O_i</math> and <math>r_i</math> be the center and radius of <math>\omega_i</math>, and let <math>O</math> and <math>r</math> be the center and radius of <math>\omega</math>. |
Since <math>\overline{AB}</math> extends to an arc with arc <math>120^\circ</math>, the distance from <math>O</math> to <math>\overline{AB}</math> is <math>r/2</math>. Let <math>X=\overline{AB}\cap \overline{O_1O_2}</math>. Consider <math>\triangle OO_1O_2</math>. The line <math>\overline{AB}</math> is perpendicular to <math>\overline{O_1O_2}</math> and passes through <math>X</math>. Let <math>H</math> be the foot from <math>O</math> to <math>\overline{O_1O_2}</math>; so <math>HX=r/2</math>. We have by tangency <math>OO_1=r+r_1</math> and <math>OO_2=r+r_2</math>. Let <math>O_1O_2=d</math>. | Since <math>\overline{AB}</math> extends to an arc with arc <math>120^\circ</math>, the distance from <math>O</math> to <math>\overline{AB}</math> is <math>r/2</math>. Let <math>X=\overline{AB}\cap \overline{O_1O_2}</math>. Consider <math>\triangle OO_1O_2</math>. The line <math>\overline{AB}</math> is perpendicular to <math>\overline{O_1O_2}</math> and passes through <math>X</math>. Let <math>H</math> be the foot from <math>O</math> to <math>\overline{O_1O_2}</math>; so <math>HX=r/2</math>. We have by tangency <math>OO_1=r+r_1</math> and <math>OO_2=r+r_2</math>. Let <math>O_1O_2=d</math>. | ||
− | + | <asy> | |
unitsize(3cm); | unitsize(3cm); | ||
pointpen=black; pointfontpen=fontsize(9); | pointpen=black; pointfontpen=fontsize(9); | ||
Line 40: | Line 40: | ||
− | + | </asy> | |
− | Since <math>X</math> is on the radical axis of <math>\omega_1</math> and <math>\omega_2</math>, it has equal power, so | + | Since <math>X</math> is on the radical axis of <math>\omega_1</math> and <math>\omega_2</math>, it has equal power with respect to both circles, so |
<cmath> O_1X^2 - r_1^2 = O_2X^2-r_2^2 \implies O_1X-O_2X = \frac{r_1^2-r_2^2}{d} </cmath>since <math>O_1X+O_2X=d</math>. Now we can solve for <math>O_1X</math> and <math>O_2X</math>, and in particular, | <cmath> O_1X^2 - r_1^2 = O_2X^2-r_2^2 \implies O_1X-O_2X = \frac{r_1^2-r_2^2}{d} </cmath>since <math>O_1X+O_2X=d</math>. Now we can solve for <math>O_1X</math> and <math>O_2X</math>, and in particular, | ||
− | \begin{align*} | + | <cmath>\begin{align*} |
O_1H &= O_1X - HX = \frac{d+\frac{r_1^2-r_2^2}{d}}{2} - \frac{r}{2} \\ | O_1H &= O_1X - HX = \frac{d+\frac{r_1^2-r_2^2}{d}}{2} - \frac{r}{2} \\ | ||
O_2H &= O_2X + HX = \frac{d-\frac{r_1^2-r_2^2}{d}}{2} + \frac{r}{2}. | O_2H &= O_2X + HX = \frac{d-\frac{r_1^2-r_2^2}{d}}{2} + \frac{r}{2}. | ||
− | \end{align*}We want to solve for <math>d</math>. By the Pythagorean Theorem (twice): | + | \end{align*}</cmath> |
− | \begin{align*} | + | We want to solve for <math>d</math>. By the Pythagorean Theorem (twice): |
− | &\qquad OH^2 = O_2H^2 - (r+r_2)^2 = O_1H^2 - (r+r_1)^2 \\ | + | <cmath>\begin{align*} |
+ | &\qquad -OH^2 = O_2H^2 - (r+r_2)^2 = O_1H^2 - (r+r_1)^2 \\ | ||
&\implies \left(d+r-\tfrac{r_1^2-r_2^2}{d}\right)^2 - 4(r+r_2)^2 = \left(d-r+\tfrac{r_1^2-r_2^2}{d}\right)^2 - 4(r+r_1)^2 \\ | &\implies \left(d+r-\tfrac{r_1^2-r_2^2}{d}\right)^2 - 4(r+r_2)^2 = \left(d-r+\tfrac{r_1^2-r_2^2}{d}\right)^2 - 4(r+r_1)^2 \\ | ||
&\implies 2dr - 2(r_1^2-r_2)^2-8rr_2-4r_2^2 = -2dr+2(r_1^2-r_2^2)-8rr_1-4r_1^2 \\ | &\implies 2dr - 2(r_1^2-r_2)^2-8rr_2-4r_2^2 = -2dr+2(r_1^2-r_2^2)-8rr_1-4r_1^2 \\ | ||
&\implies 4dr = 8rr_2-8rr_1 \\ | &\implies 4dr = 8rr_2-8rr_1 \\ | ||
− | &\implies | + | &\implies d=2r_2-2r_1 |
− | \end{align*}Therefore, <math>d=2(r_2-r_1) = 2(961-625)=\boxed{672}</math>. | + | \end{align*}</cmath> |
+ | Therefore, <math>d=2(r_2-r_1) = 2(961-625)=\boxed{672}</math>. | ||
+ | |||
+ | ==Solution 2 (Official MAA, Unedited)== | ||
+ | Denote by <math>O_1</math>, <math>O_2</math>, and <math>O</math> the centers of <math>\omega_1</math>, <math>\omega_2</math>, and <math>\omega</math>, respectively. Let <math>R_1 = 961</math> and <math>R_2 = 625</math> denote the radii of <math>\omega_1</math> and <math>\omega_2</math> respectively, <math>r</math> be the radius of <math>\omega</math>, and <math>\ell</math> the distance from <math>O</math> to the line <math>AB</math>. We claim that<cmath>\dfrac{\ell}{r} = \dfrac{R_2-R_1}{d},</cmath>where <math>d = O_1O_2</math>. This solves the problem, for then the <math>\widehat{PQ} = 120^\circ</math> condition implies <math>\tfrac{\ell}r = \cos 60^\circ = \tfrac{1}{2}</math>, and then we can solve to get <math>d = \boxed{672}</math>. | ||
+ | <asy> | ||
+ | import olympiad; | ||
+ | size(230pt); | ||
+ | defaultpen(linewidth(0.8)+fontsize(10pt)); | ||
+ | real r1 = 17, r2 = 27, d = 35, r = 18; | ||
+ | pair O1 = origin, O2 = (d,0); | ||
+ | path w1 = circle(origin,r1), w2 = circle((d,0),r2), w1p = circle(origin,r1+r), w2p = circle((d,0), r2 + r); | ||
+ | pair[] X = intersectionpoints(w1,w2), Y = intersectionpoints(w1p,w2p); | ||
+ | pair O = Y[1]; | ||
+ | path w = circle(Y[1],r); | ||
+ | pair Xp = 5 * X[1] - 4 * X[0]; | ||
+ | pair[] P = intersectionpoints(Xp--X[0],w); | ||
+ | label("$O_1$",origin,N); | ||
+ | label("$O_2$",(d,0),N); | ||
+ | label("$O$",Y[1],SW); | ||
+ | draw(origin--Y[1]--(d,0)--cycle,gray(0.6)); | ||
+ | pair T = foot(O,O1,O2), Tp = foot(O,X[0],X[1]); | ||
+ | draw(Tp--O--T^^rightanglemark(O,T,O1,60)^^rightanglemark(O,Tp,X[0],60),gray(0.6)); | ||
+ | draw(w^^w1^^w2^^P[0]--X[0]); | ||
+ | dot(Y[1]^^origin^^(d,0)); | ||
+ | label("$X$",T,N,gray(0.6)); | ||
+ | label("$Y$",foot(X[0],O1,O2),NE,gray(0.6)); | ||
+ | label("$\ell$",(O+Tp)/2,S,gray(0.6)); | ||
+ | </asy> | ||
+ | |||
+ | Denote by <math>O_1</math> and <math>O_2</math> the centers of <math>\omega_1</math> and <math>\omega_2</math> respectively. Set <math>X</math> as the projection of <math>O</math> onto <math>O_1O_2</math>, and denote by <math>Y</math> the intersection of <math>AB</math> with <math>O_1O_2</math>. Note that <math>\ell = XY</math>. Now recall that<cmath>d(O_2Y-O_1Y) = O_2Y^2 - O_1Y^2 = R_2^2 - R_1^2.</cmath>Furthermore, note that<cmath>\begin{align*}d(O_2X - O_1X) &= O_2X^2 - O_1X^2= O_2O^2 - O_1O^2 \\ &= (R_2 + r)^2 - (R_1+r)^2 = (R_2^2 - R_1^2) + 2r(R_2 - R_1).\end{align*}</cmath>Substituting the first equality into the second one and subtracting yields<cmath>2r(R_2 - R_1) = d(O_2Y - O_2X) - d(O_2X - O_1X) = 2dXY,</cmath>which rearranges to the desired. | ||
+ | |||
+ | ==Solution 3 (Inversion)== | ||
+ | WLOG assume <math>\omega</math> is a line. Note the angle condition is equivalent to the angle between <math>AB</math> and <math>\omega</math> being <math>60^\circ</math>. We claim the angle between <math>AB</math> and <math>\omega</math> is fixed as <math>\omega</math> varies. | ||
+ | |||
+ | Proof: Perform an inversion at <math>A</math>, sending <math>\omega_1</math> and <math>\omega_2</math> to two lines <math>\ell_1</math> and <math>\ell_2</math> intersecting at <math>B'</math>. Then <math>\omega</math> is sent to a circle tangent to lines <math>\ell_1</math> and <math>\ell_2</math>, which clearly intersects <math>AB'</math> at a fixed angle. Therefore the angle between <math>AB</math> and <math>\omega</math> is fixed as <math>\omega</math> varies. | ||
+ | |||
+ | Now simply take <math>\omega</math> to be a line. If <math>\omega</math> intersects <math>\omega_1</math> and <math>\omega_2</math> and <math>X,Y</math>, respectively, and the circles' centers are <math>O_1</math> and <math>O_2</math>, then the projection of <math>O_2</math> to <math>O_1X</math> at <math>F</math> gives that <math>O_2FO_1</math> is a <math>30\text{-}60\text{-}90</math> triangle. Therefore,<cmath>O_1O_2=2O_1F=2(O_1X-O_2Y)=2(961-625)=\boxed{672}.</cmath> | ||
+ | |||
+ | ~spartacle | ||
+ | |||
+ | ==Solution 4 (Radical Axis, Harmonic Quadrilaterals, and Similar Triangles)== | ||
+ | Suppose we label the points as shown [https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvNC9mLzRiM2JjYThjYmZlY2ViZGI0ODhjYzE4YzMyMmM0M2QyOTZlMmU5LmpwZw==&rn=MTU4ODUxMDg3XzczMDI0ODE4MTAwNjA5N184NDQzMjQxMjM3MDQ2NzQ5NjM4X24uanBn here]. By radical axis, the tangents to <math>\omega</math> at <math>D</math> and <math>E</math> intersect on <math>AB</math>. Thus <math>PDQE</math> is harmonic, so the tangents to <math>\omega</math> at <math>P</math> and <math>Q</math> intersect at <math>X \in DE</math>. Moreover, <math>OX \parallel O_1O_2</math> because both <math>OX</math> and <math>O_1O_2</math> are perpendicular to <math>AB</math>, and <math>OX = 2OP</math> because <math>\angle POQ = 120^{\circ}</math>. Thus<cmath>O_1O_2 = O_1Y - O_2Y = 2 \cdot 961 - 2\cdot 625 = \boxed{672}</cmath>by similar triangles. | ||
+ | |||
+ | ~mathman3880 | ||
+ | |||
+ | ==Video Solution== | ||
+ | Who wanted to see animated video solutions can see this. I found this really helpful. | ||
+ | |||
+ | https://youtu.be/YtZ8_7i833E | ||
+ | |||
+ | P.S: This video is not made by me. And solution is same like below solutions. | ||
+ | |||
+ | ≈@rounak138 | ||
==See also== | ==See also== |
Revision as of 06:28, 23 April 2021
Contents
Problem
Circles and with radii and , respectively, intersect at distinct points and . A third circle is externally tangent to both and . Suppose line intersects at two points and such that the measure of minor arc is . Find the distance between the centers of and .
Solution 1
Let and be the center and radius of , and let and be the center and radius of .
Since extends to an arc with arc , the distance from to is . Let . Consider . The line is perpendicular to and passes through . Let be the foot from to ; so . We have by tangency and . Let . Since is on the radical axis of and , it has equal power with respect to both circles, so since . Now we can solve for and , and in particular, We want to solve for . By the Pythagorean Theorem (twice): Therefore, .
Solution 2 (Official MAA, Unedited)
Denote by , , and the centers of , , and , respectively. Let and denote the radii of and respectively, be the radius of , and the distance from to the line . We claim thatwhere . This solves the problem, for then the condition implies , and then we can solve to get .
Denote by and the centers of and respectively. Set as the projection of onto , and denote by the intersection of with . Note that . Now recall thatFurthermore, note thatSubstituting the first equality into the second one and subtracting yieldswhich rearranges to the desired.
Solution 3 (Inversion)
WLOG assume is a line. Note the angle condition is equivalent to the angle between and being . We claim the angle between and is fixed as varies.
Proof: Perform an inversion at , sending and to two lines and intersecting at . Then is sent to a circle tangent to lines and , which clearly intersects at a fixed angle. Therefore the angle between and is fixed as varies.
Now simply take to be a line. If intersects and and , respectively, and the circles' centers are and , then the projection of to at gives that is a triangle. Therefore,
~spartacle
Solution 4 (Radical Axis, Harmonic Quadrilaterals, and Similar Triangles)
Suppose we label the points as shown here. By radical axis, the tangents to at and intersect on . Thus is harmonic, so the tangents to at and intersect at . Moreover, because both and are perpendicular to , and because . Thusby similar triangles.
~mathman3880
Video Solution
Who wanted to see animated video solutions can see this. I found this really helpful.
P.S: This video is not made by me. And solution is same like below solutions.
≈@rounak138
See also
2021 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.