Difference between revisions of "2021 AIME I Problems/Problem 13"
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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 | + | &\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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==Solution 2 (Linearity)== | ==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 | + | 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== | ==Solution 3== | ||
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==Solution 4 (Quick)== | ==Solution 4 (Quick)== | ||
− | Suppose we label the points as shown [https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvNC9mLzRiM2JjYThjYmZlY2ViZGI0ODhjYzE4YzMyMmM0M2QyOTZlMmU5LmpwZw==&rn=MTU4ODUxMDg3XzczMDI0ODE4MTAwNjA5N184NDQzMjQxMjM3MDQ2NzQ5NjM4X24uanBn | + | 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; | ||
+ | |||
+ | 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 <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 | ||
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'''vladimir.shelomovskii@gmail.com, vvsss''' | '''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> | ||
+ | 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: | ||
+ | <cmath>h^2+r^2 = 625^2 \; (2)</cmath> | ||
+ | <cmath>h^2+(x-r)^2 = 961^2 \; (3)</cmath> | ||
+ | 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>. | ||
+ | |||
+ | ~sml1809 | ||
+ | |||
+ | ==Solution 8 (Cheese)== | ||
+ | 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>. | ||
+ | |||
+ | Notice the 30-60-90 triangle in the image. <math>O_1R = 961 - 625</math>. | ||
+ | |||
+ | Thus, the distance between the centers of <math>\omega_1</math> and <math>\omega_2</math> is <math>2(961-625)=\boxed{672}</math> | ||
+ | |||
+ | [[File:Cheesed circle.png|thumb|800px]] | ||
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+ | picture by afly | ||
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+ | ==Video Solution by MOP 2024== | ||
+ | https://youtu.be/GQT73xqvtXA | ||
+ | |||
+ | ~r00tsOfUnity | ||
==Video Solution== | ==Video Solution== | ||
− | https://youtu.be/gN7Ocu3D62M | + | https://youtu.be/gN7Ocu3D62M |
+ | ~Math Problem Solving Skills | ||
==Video Solution== | ==Video Solution== |
Latest revision as of 03:59, 7 February 2024
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
.
Quick, Olympiad-Style Solution
Let and
. The solution relies on the following key claim:
Claim. is a harmonic quadrilateral.
Proof. Using the radical axis theorem, the tangents to circle at
and
are concurrent with line
at the radical center, implying that our claim is true by harmonic quadrilateral properties.
Thus, we deduce the tangents to at
and
are concurrent at
with line
. Denote by
the centers of
respectively.
Now suppose . Note that
giving us the pairs of similar triangles
We thereby obtain
since
, where
denotes the radius of
, and
as well. It follows that
Solution 1 (Radical Axis)
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 (Linearity)
Let ,
, and
be the centers of
,
, and
with
,
, and
their radii, respectively. Then, the distance from
to the radical axis
of
is equal to
. Let
and
the orthogonal projection of
onto line
. Define the function
by
Then
By linearity,
Notice that
and
, thus
Dividing both sides by
(which is obviously nonzero as
is nonzero) gives us
so
. Since
and
, the answer is
.
Solution 3
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 that
where
. 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 that
Furthermore, note that
Substituting the first equality into the second one and subtracting yields
which rearranges to the desired.
Solution 4 (Quick)
Suppose we label the points as shown below.
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
. Thus
by similar triangles.
~mathman3880
Solution 5 (Official MAA)
Like in other solutions, let be the center of
with
its radius; also, let
and
be the centers of
and
with
and
their radii, respectively. Let line
intersect line
at
, and let
,
,
, where the length
splits as
. Because the lines
and
are perpendicular, lines
and
meet at a
angle.
Applying the Law of Cosines to ,
,
, and
gives
Adding the first and fourth equations, then subtracting the second and third equations gives us
Since lies on the radical axis of
and
, the power of point
with respect to either circle is
Hence which simplifies to
The requested distance is therefore equal to
.
Solution 6 (Geometry)
Let circle tangent circles
and
respectively at distinct points
and
. Let
be the centers (the radii) of
and
respectively. WLOG
Let
be the point of
such, that
Let
be the midpoint
be the radical axes of
and
Then is radical center of
and
In is bisector of
is median
and
are collinear.
is cyclic (in
is diameter
as they intercept the same arc
in
Since and
, the answer is
.
vladimir.shelomovskii@gmail.com, vvsss
Solution 7
We are not given the radius of circle , but based on the problem statement, that radius isn't important. We can set
to have radius infinity (solution 8), but if you didn't observe that, you could also set the radius to be
so that the line containing the center of
, call it
, and
, call it
, is perpendicular to the line containing the center of
, call it
and
. Let
and
. Also, let the projections of
and
onto line
be
and
, respectively.
By Pythagorean Theorem on , we get
Note that since
,
. So,
. We now get two more equations from Pythag:
From subtracting
and
,
. Rearranging
yields
. Plugging in our result from
,
.
~sml1809
Solution 8 (Cheese)
Let the circle be infinitely big (a line). Then for it to be split into an arc of
,
must intersect at a
with line
.
Notice the 30-60-90 triangle in the image. .
Thus, the distance between the centers of and
is
picture by afly
Video Solution by MOP 2024
~r00tsOfUnity
Video Solution
~Math Problem Solving Skills
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.