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Difference between revisions of "2021 AIME I Problems/Problem 13"

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(→‎Solution 1 (Radical Axis))
 
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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>.
 
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 1==
+
==Quick, Olympiad-Style Solution==
 +
 
 +
Let <math>D\equiv\omega\cap\omega_{1}</math> and <math>E\equiv\omega\cap\omega_{2}</math>. The solution relies on the following key claim:
 +
 
 +
<font color=red><b>Claim.</b></font color> <math>DPEQ</math> is a harmonic quadrilateral.
 +
 
 +
<font color=blue><b>Proof.</b></font color> Using the radical axis theorem, the tangents to circle <math>\omega</math> at <math>D</math> and <math>E</math> are concurrent with line <math>\overline{ABPQ}</math> at the radical center, implying that our claim is true by harmonic quadrilateral properties. <math>\blacksquare</math>
 +
 
 +
Thus, we deduce the tangents to <math>\omega</math> at <math>P</math> and <math>Q</math> are concurrent at <math>F</math> with line <math>\overline{DE}</math>. Denote by <math>O_{1}, O_{2}, O</math> the centers of <math>\omega_{1}, \omega_{2}, \omega</math> respectively.
 +
 
 +
Now suppose <math>G\equiv\overline{FDE}\cap\overline{O_{1}O_{2}}</math>. Note that <math>\overline{OF}\parallel\overline{O_{1}O_{2}}</math> giving us the pairs of similar triangles <cmath>\triangle O_{1}GD\sim\triangle OFD~\text{and}~\triangle O_{2}GE\sim\triangle OFE.</cmath> We thereby obtain <cmath>\dfrac{O_{1}G}{O_{1}D}=\dfrac{OF}{OD}=\dfrac{r\sec\tfrac{120^{\circ}}{2}}{r}=\sec 60^{\circ}=2</cmath> since <math>\widehat{PQ}=120^{\circ}</math>, where <math>r</math> denotes the radius of <math>\omega</math>, and <cmath>\dfrac{O_{2}G}{O_{2}E}=\dfrac{OF}{OE}=\sec 60^{\circ}=2</cmath> as well. It follows that <cmath>O_{1}O_{2}=O_{1}G-O_{2}G=2(O_{1}D-O_{2}E)=2(961-625)=\textbf{672}.</cmath>
 +
 
 +
[[File:AIME 2021-I13 Geogebra Diagram.png|672px]]
 +
 
 +
==Solution 1 (Radical Axis)==
 
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>.
 
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>.
  
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&\qquad -OH^2 = O_2H^2 - (r+r_2)^2 = O_1H^2 - (r+r_1)^2 \\
 
&\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 d=2r_2-2r_1
 
&\implies d=2r_2-2r_1
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Therefore, <math>d=2(r_2-r_1) = 2(961-625)=\boxed{672}</math>.
 
Therefore, <math>d=2(r_2-r_1) = 2(961-625)=\boxed{672}</math>.
  
−
==Solution 2 (Official MAA, Unedited)==
+
==Solution 2 (Linearity)==
 +
Let <math>O_{1}</math>, <math>O_{2}</math>, and <math>O</math> be the centers of <math>\omega_{1}</math>, <math>\omega_{2}</math>, and <math>\omega</math> with <math>r_{1}</math>, <math>r_{2}</math>, and <math>r</math> their radii, respectively. Then, the distance from <math>O</math> to the radical axis <math>\ell\equiv\overline{AB}</math> of <math>\omega_{1}, \omega_{2}</math> is equal to <math>\frac{1}{2}r</math>. Let <math>x=O_{1}O_{2}</math> and <math>O^{\prime}</math> the orthogonal projection of <math>O</math> onto line <math>\ell</math>. Define the function <math>f:\mathbb{R}^{2}\rightarrow\mathbb{R}</math> by <cmath>f(X)=\text{Pow}_{\omega_{1}}(X)-\text{Pow}_{\omega_{2}}(X).</cmath> Then <cmath>\begin{align*} f(O_{1})=-r_{1}^{2}-(x-r_{2})(x+r_{2})&=-x^{2}+r_{2}^{2}-r_{1}^{2}, \\ f(O_{2})=(x-r_{1})(x+r_{1})-(-r_{2}^{2})&=x^{2}+r_{2}^{2}-r_{1}^{2}, \\ f(O)=r(r+2r_{1})-r(r+2r_{2})&=2r(r_{1}-r_{2}), \\ f(O^{\prime})&=0. \end{align*}</cmath> By [https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvMC84LzkzZjZjZmFlMGViY2E3MDMxNWQzY2IzNzFlZTk5NWFmOTM5ZGY1LnBkZg==&rn=TGluZWFyaXR5IG9mIFBvd2VyIG9mIGEgUG9pbnQucGRm linearity], <cmath>\frac{f(O_{2})-f(O_{1})}{f(O)-f(O^{\prime})}=\frac{O_{1}O_{2}}{OO^{\prime}}=\frac{x}{\tfrac{1}{2}r}=\frac{2x}{r}.</cmath> Notice that <math>f(O_{2})-f(O_{1})=x^{2}-(-x^{2})=2x^{2}</math> and <math>f(O)-f(O^{\prime})=2r(r_{1}-r_{2})</math>, thus <cmath>\begin{align*}\frac{2x^{2}}{2r(r_{1}-r_{2})}&=\frac{2x}{r}\end{align*}</cmath> Dividing both sides by <math>\frac{2x}{r}</math> (which is obviously nonzero as <math>x</math> is nonzero) gives us <cmath>\begin{align*}\frac{x}{2(r_{1}-r_{2})}&=1\end{align*}</cmath> so <math>x=2(r_{1}-r_{2})</math>. Since <math>r_{1}=961</math> and <math>r_{2}=625</math>, the answer is <math>x=2\cdot(961-625)=\boxed{672}</math>.
 +
 
 +
==Solution 3==
 
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>.
 
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>
 
<asy>
Line 84: Line 101:
 
</asy>
 
</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.
+
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_2X - O_1X) - d(O_2Y - O_1Y) = 2dXY,</cmath>which rearranges to the desired.
  
−
==Solution 3 (Inversion)==
+
==Solution 4 (Quick)==
−
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.
+
Suppose we label the points as shown [https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvNC9mLzRiM2JjYThjYmZlY2ViZGI0ODhjYzE4YzMyMmM0M2QyOTZlMmU5LmpwZw==&rn=MTU4ODUxMDg3XzczMDI0ODE4MTAwNjA5N184NDQzMjQxMjM3MDQ2NzQ5NjM4X24uanBn below].  
 +
<asy>
 +
defaultpen(fontsize(12)+0.6); size(300);
 +
pen p=fontsize(10)+royalblue+0.4;
  
−
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.
+
var r=1200;
 +
pair O1=origin, O2=(672,0), O=OP(CR(O1,961+r),CR(O2,625+r));
 +
path c1=CR(O1,961), c2=CR(O2,625), c=CR(O,r);
 +
pair
 +
A=IP(CR(O1,961),CR(O2,625)), B=OP(CR(O1,961),CR(O2,625)),
 +
P=IP(L(A,B,0,0.2),c),
 +
Q=IP(L(A,B,0,200),c),
 +
F=IP(CR(O,625+r),O--O1),
 +
M=(F+O2)/2,
 +
D=IP(CR(O,r),O--O1),
 +
E=IP(CR(O,r),O--O2),
 +
X=extension(E,D,O,O+O1-O2),
 +
Y=extension(D,E,O1,O2);
  
−
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>
+
draw(c1^^c2); draw(c,blue+0.6); draw(O1--O2--O--cycle,black+0.6); draw(O--X^^Y--O2,black+0.6); draw(X--Y,heavygreen+0.6);
 +
draw((X+O)/2--O,MidArrow); draw(O2--Y-(300,0),MidArrow);
  
−
~spartacle
+
dot("$A$",A,dir(A-O2/2));
 +
dot("$B$",B,dir(B-O2/2));
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dot("$O_2$",O2,right+up);  dot("$O_1$",O1,left+up);
 +
dot("$O$",O,dir(O-O2));
 +
dot("$D$",D,dir(170));
 +
dot("$E$",E,dir(E-O1));
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dot("$X$",X,dir(X-D));
 +
dot("$Y$",Y,dir(Y-D));
 +
 +
label("$R$",O--E,right+up,p);
 +
label("$R$",O--D,left+down,p);
 +
label("$2R$",(X+O)/2-(150,0),down,p);
 +
label("$961$",O1--D,2*(left+down),p);
 +
label("$625$",O2--E,2*(right+up),p);
  
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==Solution 4 (Radical Axis, Harmonic Quadrilaterals, and Similar Triangles)==
+
MA("",E,D,O1,100,fuchsia+linewidth(1));
−
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.
+
MA("",X,D,O,100,fuchsia+linewidth(1));
 +
MA("",Y,E,O2,100,orange+linewidth(1));
 +
MA("",D,E,O,100,orange+linewidth(1));
 +
</asy>
 +
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
 
~mathman3880
 +
 +
==Solution 5 (Official MAA)==
 +
Like in other solutions, let <math>O</math> be the center of <math>\omega</math> with <math>r</math> its radius; also, let <math>O_{1}</math> and <math>O_{2}</math> be the centers of <math>\omega_{1}</math> and <math>\omega_{2}</math> with <math>R_{1}</math> and <math>R_{2}</math> their radii, respectively. Let line <math>OP</math> intersect line <math>O_{1}O_{2}</math> at <math>T</math>, and let <math>u=TO_{2}</math>, <math>v=TO_{1}</math>, <math>x=PT</math>, where the length <math>O_{1}O_{2}</math> splits as <math>u+v</math>. Because the lines <math>PQ</math> and <math>O_{1}O_{2}</math> are perpendicular, lines <math>OT</math> and <math>O_{1}O_{2}</math> meet at a <math>60^{\circ}</math> angle.
 +
 +
Applying the Law of Cosines to <math>\triangle O_{2}PT</math>, <math>\triangle O_{1}PT</math>, <math>\triangle O_{2}OT</math>, and <math>\triangle O_{1}OT</math> gives <cmath>\begin{align*}\triangle O_{2}PT&:O_{2}P^{2}=u^{2}+x^{2}-ux \\ \triangle O_{1}PT&:O_{1}P^{2}=v^{2}+x^{2}+vx \\ \triangle O_{2}OT&:(r+R_{2})^{2}=u^{2}+(r+x)^{2}-u(r+x) \\ \triangle O_{1}OT&:(r+R_{1})^{2}=v^{2}+(r+x)^{2}+v(r+x)\end{align*}</cmath>
 +
 +
Adding the first and fourth equations, then subtracting the second and third equations gives us <cmath>\left(O_{2}P^{2}-O_{1}P^{2}\right)+\left(R_{1}^{2}-R_{2}^{2}\right)+2r(R_{1}-R_{2})=r(u+v)</cmath>
 +
 +
Since <math>P</math> lies on the radical axis of <math>\omega_{1}</math> and <math>\omega_{2}</math>, the power of point <math>P</math> with respect to either circle is <cmath>O_{2}P^{2}-R_{2}^{2}=O_{1}P^{2}-R_{1}^{2}.</cmath>
 +
 +
Hence <math>2r(R_{1}-R_{2})=r(u+v)</math> which simplifies to <cmath>u+v=2(R_{1}-R_{2}).</cmath>
 +
 +
The requested distance <cmath>O_{1}O_{2}=O_{1}T+O_{2}T=u+v</cmath> is therefore equal to <math>2\cdot(961-625)=\boxed{672}</math>.
 +
 +
==Solution 6 (Geometry)==
 +
[[File:2021 AIME I 13.png|500px|right]]
 +
Let circle <math>\omega</math> tangent circles <math>\omega_1</math> and <math>\omega_2,</math> respectively at distinct points <math>C</math> and <math>D</math>. Let <math>O, O_1, O_2 (r, r_1, r_2)</math> be the centers (the radii) of <math>\omega, \omega_1</math> and <math>\omega_2,</math> respectively. WLOG <math>r_1 < r_2.</math> Let <math>F</math> be the point of <math>OO_2</math> such, that <math>OO_1 =OF.</math> Let <math>M</math> be the midpoint <math>FO_1, OE \perp AB, CT</math> be the radical axes of  <math>\omega_1</math> and <math>\omega, T \in AB.</math>
 +
 +
Then <math>T</math> is radical center of  <math>\omega, \omega_1</math> and <math>\omega_2, TD = CT.</math>
 +
 +
In <math>\triangle OFO_1 (OF = OO_1) OT</math> is bisector of <math>\angle O, OM</math> is median
 +
 +
<math>\hspace{10mm} \implies O,T,</math>  and <math>M</math> are collinear.
 +
 +
<cmath>\angle OCT = \angle ODT = \angle OET = 90^\circ \implies</cmath>
 +
 +
<math>OCTDE</math> is cyclic (in <math>\Omega), OT</math> is diameter <math>\Omega.</math>
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<math>O_1O_2 \perp AB, OM \perp FO_1 \implies \angle FO_1O_2 = \angle OTE</math>
 +
<math>\angle OTE = \angle ODE</math> as they intercept the same arc  <math> \overset{\Large\frown}{OE}</math> in <math>\Omega.</math>
 +
<cmath>OE \perp AB, O_1O_2 \perp AB \implies O_1 O_2 || OE \implies</cmath>
 +
<cmath>\angle OO_2O_1 = \angle O_2 OE \implies \triangle ODE \sim \triangle O_2 O_1 F \implies</cmath>
 +
<cmath>\frac {OE}{OD} = \frac {O_2F}{O_1O_2} \implies \cos \frac {120^\circ}{2} = \frac{r_2 + r - r_1 -r} {O_1O_2}\implies {O_1O_2}= 2|r_2 – r_1|.</cmath>
 +
 +
Since <math>r_{1}=625</math> and <math>r_{2}=961</math>, the answer is <math>2\cdot|961-625|=\boxed{672}</math>.
 +
 +
'''vladimir.shelomovskii@gmail.com, vvsss'''
 +
 +
==Solution 7==
 +
We are not given the radius of circle <math>w</math>, but based on the problem statement, that radius isn't important. We can set <math>w</math> to have radius infinity (solution 8), but if you didn't observe that, you could also set the radius to be <math>2r</math> so that the line containing the center of <math>w</math>, call it <math>W</math>, and <math>w_2</math>, call it <math>W_2</math>, is perpendicular to the line containing the center of <math>w_1</math>, call it <math>W_1</math> and <math>w_2</math>. Let <math>AB = 2h</math> and <math>W_1W_2 = x</math>. Also, let the projections of <math>W</math> and <math>W_1</math> onto line <math>AB</math> be <math>X</math> and <math>Y</math>, respectively.
 +
 +
By [[Pythagorean Theorem]] on <math>\triangle WW_1W_2</math>, we get
 +
<cmath>x^2+(625+2x)^2=(961+2r)^2 \;(1)</cmath>
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Note that since <math>\angle PWQ = 120</math>, <math>\angle PWX = 60</math>. So, <math>WX = 2r/2 = r = W_1Y</math>. We now get two more equations from Pythag:
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<cmath>h^2+r^2 = 625^2 \; (2)</cmath>
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<cmath>h^2+(x-r)^2 = 961^2 \; (3)</cmath>
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From subtracting <math>(2)</math> and <math>(3)</math>, <math>x^2-2rx=961^2-625^2 \; (4)</math>. Rearranging <math>(3)</math> yields <math>x^2-1344r = 961^2+625^2</math>. Plugging in our result from <math>(4)</math>, <math>x^2-1344r= x^2-2rx \implies 1344r = 2rx \implies x=\boxed{672}</math>.
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~sml1809
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==Solution 8 (Cheese)==
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Let the [[circle]] <math>\omega</math> be infinitely big (a line). Then for it to be split into an [[arc]] of <math>120^{\circ}</math>, <math>\overline{PQ}</math> must intersect at a <math>60^{\circ}</math> with line <math>\omega</math>.
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Notice the 30-60-90 triangle in the image. <math>O_1R = 961 - 625</math>.
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Thus, the distance between the centers of <math>\omega_1</math> and <math>\omega_2</math> is <math>2(961-625)=\boxed{672}</math>
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[[File:Cheesed circle.png|thumb|800px]]
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picture by afly
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==Video Solution by MOP 2024==
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https://youtu.be/GQT73xqvtXA
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~r00tsOfUnity
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==Video Solution==
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https://youtu.be/gN7Ocu3D62M
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~Math Problem Solving Skills
  
 
==Video Solution==
 
==Video Solution==

Latest revision as of 02:59, 7 February 2024

Problem

Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$, respectively, intersect at distinct points $A$ and $B$. A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$. Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $120^{\circ}$. Find the distance between the centers of $\omega_1$ and $\omega_2$.

Quick, Olympiad-Style Solution

Let $D\equiv\omega\cap\omega_{1}$ and $E\equiv\omega\cap\omega_{2}$. The solution relies on the following key claim:

Claim. $DPEQ$ is a harmonic quadrilateral.

Proof. Using the radical axis theorem, the tangents to circle $\omega$ at $D$ and $E$ are concurrent with line $\overline{ABPQ}$ at the radical center, implying that our claim is true by harmonic quadrilateral properties. $\blacksquare$

Thus, we deduce the tangents to $\omega$ at $P$ and $Q$ are concurrent at $F$ with line $\overline{DE}$. Denote by $O_{1}, O_{2}, O$ the centers of $\omega_{1}, \omega_{2}, \omega$ respectively.

Now suppose $G\equiv\overline{FDE}\cap\overline{O_{1}O_{2}}$. Note that $\overline{OF}\parallel\overline{O_{1}O_{2}}$ giving us the pairs of similar triangles \[\triangle O_{1}GD\sim\triangle OFD~\text{and}~\triangle O_{2}GE\sim\triangle OFE.\] We thereby obtain \[\dfrac{O_{1}G}{O_{1}D}=\dfrac{OF}{OD}=\dfrac{r\sec\tfrac{120^{\circ}}{2}}{r}=\sec 60^{\circ}=2\] since $\widehat{PQ}=120^{\circ}$, where $r$ denotes the radius of $\omega$, and \[\dfrac{O_{2}G}{O_{2}E}=\dfrac{OF}{OE}=\sec 60^{\circ}=2\] as well. It follows that \[O_{1}O_{2}=O_{1}G-O_{2}G=2(O_{1}D-O_{2}E)=2(961-625)=\textbf{672}.\]

AIME 2021-I13 Geogebra Diagram.png

Solution 1 (Radical Axis)

Let $O_i$ and $r_i$ be the center and radius of $\omega_i$, and let $O$ and $r$ be the center and radius of $\omega$.

Since $\overline{AB}$ extends to an arc with arc $120^\circ$, the distance from $O$ to $\overline{AB}$ is $r/2$. Let $X=\overline{AB}\cap \overline{O_1O_2}$. Consider $\triangle OO_1O_2$. The line $\overline{AB}$ is perpendicular to $\overline{O_1O_2}$ and passes through $X$. Let $H$ be the foot from $O$ to $\overline{O_1O_2}$; so $HX=r/2$. We have by tangency $OO_1=r+r_1$ and $OO_2=r+r_2$. Let $O_1O_2=d$. [asy] unitsize(3cm); pointpen=black; pointfontpen=fontsize(9); pair A=dir(110), B=dir(230), C=dir(310); DPA(A--B--C--A); pair H = foot(A, B, C); draw(A--H); pair X = 0.3*B + 0.7*C; pair Y = A+X-H; draw(X--1.3*Y-0.3*X); draw(A--Y, dotted); pair R1 = 1.3*X-0.3*Y; pair R2 = 0.7*X+0.3*Y; draw(R1--X); D("O",A,dir(A)); D("O_1",B,dir(B)); D("O_2",C,dir(C)); D("H",H,dir(270)); D("X",X,dir(225)); D("A",R1,dir(180)); D("B",R2,dir(180)); draw(rightanglemark(Y,X,C,3)); [/asy] Since $X$ is on the radical axis of $\omega_1$ and $\omega_2$, it has equal power with respect to both circles, so \[O_1X^2 - r_1^2 = O_2X^2-r_2^2 \implies O_1X-O_2X = \frac{r_1^2-r_2^2}{d}\]since $O_1X+O_2X=d$. Now we can solve for $O_1X$ and $O_2X$, and in particular, \begin{align*} 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}. \end{align*} We want to solve for $d$. By the Pythagorean Theorem (twice): \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 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 d=2r_2-2r_1 \end{align*} Therefore, $d=2(r_2-r_1) = 2(961-625)=\boxed{672}$.

Solution 2 (Linearity)

Let $O_{1}$, $O_{2}$, and $O$ be the centers of $\omega_{1}$, $\omega_{2}$, and $\omega$ with $r_{1}$, $r_{2}$, and $r$ their radii, respectively. Then, the distance from $O$ to the radical axis $\ell\equiv\overline{AB}$ of $\omega_{1}, \omega_{2}$ is equal to $\frac{1}{2}r$. Let $x=O_{1}O_{2}$ and $O^{\prime}$ the orthogonal projection of $O$ onto line $\ell$. Define the function $f:\mathbb{R}^{2}\rightarrow\mathbb{R}$ by \[f(X)=\text{Pow}_{\omega_{1}}(X)-\text{Pow}_{\omega_{2}}(X).\] Then \begin{align*} f(O_{1})=-r_{1}^{2}-(x-r_{2})(x+r_{2})&=-x^{2}+r_{2}^{2}-r_{1}^{2}, \\ f(O_{2})=(x-r_{1})(x+r_{1})-(-r_{2}^{2})&=x^{2}+r_{2}^{2}-r_{1}^{2}, \\ f(O)=r(r+2r_{1})-r(r+2r_{2})&=2r(r_{1}-r_{2}), \\ f(O^{\prime})&=0. \end{align*} By linearity, \[\frac{f(O_{2})-f(O_{1})}{f(O)-f(O^{\prime})}=\frac{O_{1}O_{2}}{OO^{\prime}}=\frac{x}{\tfrac{1}{2}r}=\frac{2x}{r}.\] Notice that $f(O_{2})-f(O_{1})=x^{2}-(-x^{2})=2x^{2}$ and $f(O)-f(O^{\prime})=2r(r_{1}-r_{2})$, thus \begin{align*}\frac{2x^{2}}{2r(r_{1}-r_{2})}&=\frac{2x}{r}\end{align*} Dividing both sides by $\frac{2x}{r}$ (which is obviously nonzero as $x$ is nonzero) gives us \begin{align*}\frac{x}{2(r_{1}-r_{2})}&=1\end{align*} so $x=2(r_{1}-r_{2})$. Since $r_{1}=961$ and $r_{2}=625$, the answer is $x=2\cdot(961-625)=\boxed{672}$.

Solution 3

Denote by $O_1$, $O_2$, and $O$ the centers of $\omega_1$, $\omega_2$, and $\omega$, respectively. Let $R_1 = 961$ and $R_2 = 625$ denote the radii of $\omega_1$ and $\omega_2$ respectively, $r$ be the radius of $\omega$, and $\ell$ the distance from $O$ to the line $AB$. We claim that\[\dfrac{\ell}{r} = \dfrac{R_2-R_1}{d},\]where $d = O_1O_2$. This solves the problem, for then the $\widehat{PQ} = 120^\circ$ condition implies $\tfrac{\ell}r = \cos 60^\circ = \tfrac{1}{2}$, and then we can solve to get $d = \boxed{672}$. [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 $O_1$ and $O_2$ the centers of $\omega_1$ and $\omega_2$ respectively. Set $X$ as the projection of $O$ onto $O_1O_2$, and denote by $Y$ the intersection of $AB$ with $O_1O_2$. Note that $\ell = XY$. Now recall that\[d(O_2Y-O_1Y) = O_2Y^2 - O_1Y^2 = R_2^2 - R_1^2.\]Furthermore, note that\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*}Substituting the first equality into the second one and subtracting yields\[2r(R_2 - R_1) = d(O_2X - O_1X) - d(O_2Y - O_1Y) = 2dXY,\]which rearranges to the desired.

Solution 4 (Quick)

Suppose we label the points as shown below. [asy] defaultpen(fontsize(12)+0.6); size(300); pen p=fontsize(10)+royalblue+0.4; var r=1200; pair O1=origin, O2=(672,0), O=OP(CR(O1,961+r),CR(O2,625+r)); path c1=CR(O1,961), c2=CR(O2,625), c=CR(O,r); pair A=IP(CR(O1,961),CR(O2,625)), B=OP(CR(O1,961),CR(O2,625)), P=IP(L(A,B,0,0.2),c), Q=IP(L(A,B,0,200),c), F=IP(CR(O,625+r),O--O1), M=(F+O2)/2, D=IP(CR(O,r),O--O1), E=IP(CR(O,r),O--O2), X=extension(E,D,O,O+O1-O2), Y=extension(D,E,O1,O2); draw(c1^^c2); draw(c,blue+0.6); draw(O1--O2--O--cycle,black+0.6); draw(O--X^^Y--O2,black+0.6); draw(X--Y,heavygreen+0.6); draw((X+O)/2--O,MidArrow); draw(O2--Y-(300,0),MidArrow); dot("$A$",A,dir(A-O2/2)); dot("$B$",B,dir(B-O2/2)); dot("$O_2$",O2,right+up); dot("$O_1$",O1,left+up); dot("$O$",O,dir(O-O2)); dot("$D$",D,dir(170)); dot("$E$",E,dir(E-O1)); dot("$X$",X,dir(X-D)); dot("$Y$",Y,dir(Y-D)); label("$R$",O--E,right+up,p); label("$R$",O--D,left+down,p); label("$2R$",(X+O)/2-(150,0),down,p); label("$961$",O1--D,2*(left+down),p); label("$625$",O2--E,2*(right+up),p); MA("",E,D,O1,100,fuchsia+linewidth(1)); MA("",X,D,O,100,fuchsia+linewidth(1)); MA("",Y,E,O2,100,orange+linewidth(1)); MA("",D,E,O,100,orange+linewidth(1)); [/asy] By radical axis, the tangents to $\omega$ at $D$ and $E$ intersect on $AB$. Thus $PDQE$ is harmonic, so the tangents to $\omega$ at $P$ and $Q$ intersect at $X \in DE$. Moreover, $OX \parallel O_1O_2$ because both $OX$ and $O_1O_2$ are perpendicular to $AB$, and $OX = 2OP$ because $\angle POQ = 120^{\circ}$. Thus\[O_1O_2 = O_1Y - O_2Y = 2 \cdot 961 - 2\cdot 625 = \boxed{672}\]by similar triangles.

~mathman3880

Solution 5 (Official MAA)

Like in other solutions, let $O$ be the center of $\omega$ with $r$ its radius; also, let $O_{1}$ and $O_{2}$ be the centers of $\omega_{1}$ and $\omega_{2}$ with $R_{1}$ and $R_{2}$ their radii, respectively. Let line $OP$ intersect line $O_{1}O_{2}$ at $T$, and let $u=TO_{2}$, $v=TO_{1}$, $x=PT$, where the length $O_{1}O_{2}$ splits as $u+v$. Because the lines $PQ$ and $O_{1}O_{2}$ are perpendicular, lines $OT$ and $O_{1}O_{2}$ meet at a $60^{\circ}$ angle.

Applying the Law of Cosines to $\triangle O_{2}PT$, $\triangle O_{1}PT$, $\triangle O_{2}OT$, and $\triangle O_{1}OT$ gives \begin{align*}\triangle O_{2}PT&:O_{2}P^{2}=u^{2}+x^{2}-ux \\ \triangle O_{1}PT&:O_{1}P^{2}=v^{2}+x^{2}+vx \\ \triangle O_{2}OT&:(r+R_{2})^{2}=u^{2}+(r+x)^{2}-u(r+x) \\ \triangle O_{1}OT&:(r+R_{1})^{2}=v^{2}+(r+x)^{2}+v(r+x)\end{align*}

Adding the first and fourth equations, then subtracting the second and third equations gives us \[\left(O_{2}P^{2}-O_{1}P^{2}\right)+\left(R_{1}^{2}-R_{2}^{2}\right)+2r(R_{1}-R_{2})=r(u+v)\]

Since $P$ lies on the radical axis of $\omega_{1}$ and $\omega_{2}$, the power of point $P$ with respect to either circle is \[O_{2}P^{2}-R_{2}^{2}=O_{1}P^{2}-R_{1}^{2}.\]

Hence $2r(R_{1}-R_{2})=r(u+v)$ which simplifies to \[u+v=2(R_{1}-R_{2}).\]

The requested distance \[O_{1}O_{2}=O_{1}T+O_{2}T=u+v\] is therefore equal to $2\cdot(961-625)=\boxed{672}$.

Solution 6 (Geometry)

2021 AIME I 13.png

Let circle $\omega$ tangent circles $\omega_1$ and $\omega_2,$ respectively at distinct points $C$ and $D$. Let $O, O_1, O_2 (r, r_1, r_2)$ be the centers (the radii) of $\omega, \omega_1$ and $\omega_2,$ respectively. WLOG $r_1 < r_2.$ Let $F$ be the point of $OO_2$ such, that $OO_1 =OF.$ Let $M$ be the midpoint $FO_1, OE \perp AB, CT$ be the radical axes of $\omega_1$ and $\omega, T \in AB.$

Then $T$ is radical center of $\omega, \omega_1$ and $\omega_2, TD = CT.$

In $\triangle OFO_1 (OF = OO_1) OT$ is bisector of $\angle O, OM$ is median

$\hspace{10mm} \implies O,T,$ and $M$ are collinear.

\[\angle OCT = \angle ODT = \angle OET = 90^\circ \implies\]

$OCTDE$ is cyclic (in $\Omega), OT$ is diameter $\Omega.$ $O_1O_2 \perp AB, OM \perp FO_1 \implies \angle FO_1O_2 = \angle OTE$ $\angle OTE = \angle ODE$ as they intercept the same arc $\overset{\Large\frown}{OE}$ in $\Omega.$ \[OE \perp AB, O_1O_2 \perp AB \implies O_1 O_2 || OE \implies\] \[\angle OO_2O_1 = \angle O_2 OE \implies \triangle ODE \sim \triangle O_2 O_1 F \implies\] \[\frac {OE}{OD} = \frac {O_2F}{O_1O_2} \implies \cos \frac {120^\circ}{2} = \frac{r_2 + r - r_1 -r} {O_1O_2}\implies {O_1O_2}= 2|r_2 – r_1|.\]

Since $r_{1}=625$ and $r_{2}=961$, the answer is $2\cdot|961-625|=\boxed{672}$.

vladimir.shelomovskii@gmail.com, vvsss

Solution 7

We are not given the radius of circle $w$, but based on the problem statement, that radius isn't important. We can set $w$ to have radius infinity (solution 8), but if you didn't observe that, you could also set the radius to be $2r$ so that the line containing the center of $w$, call it $W$, and $w_2$, call it $W_2$, is perpendicular to the line containing the center of $w_1$, call it $W_1$ and $w_2$. Let $AB = 2h$ and $W_1W_2 = x$. Also, let the projections of $W$ and $W_1$ onto line $AB$ be $X$ and $Y$, respectively.

By Pythagorean Theorem on $\triangle WW_1W_2$, we get \[x^2+(625+2x)^2=(961+2r)^2 \;(1)\] Note that since $\angle PWQ = 120$, $\angle PWX = 60$. So, $WX = 2r/2 = r = W_1Y$. We now get two more equations from Pythag: \[h^2+r^2 = 625^2 \; (2)\] \[h^2+(x-r)^2 = 961^2 \; (3)\] From subtracting $(2)$ and $(3)$, $x^2-2rx=961^2-625^2 \; (4)$. Rearranging $(3)$ yields $x^2-1344r = 961^2+625^2$. Plugging in our result from $(4)$, $x^2-1344r= x^2-2rx \implies 1344r = 2rx \implies x=\boxed{672}$.

~sml1809

Solution 8 (Cheese)

Let the circle $\omega$ be infinitely big (a line). Then for it to be split into an arc of $120^{\circ}$, $\overline{PQ}$ must intersect at a $60^{\circ}$ with line $\omega$.

Notice the 30-60-90 triangle in the image. $O_1R = 961 - 625$.

Thus, the distance between the centers of $\omega_1$ and $\omega_2$ is $2(961-625)=\boxed{672}$

Cheesed circle.png























picture by afly

Video Solution by MOP 2024

https://youtu.be/GQT73xqvtXA

~r00tsOfUnity

Video Solution

https://youtu.be/gN7Ocu3D62M

~Math Problem Solving Skills

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

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

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