Difference between revisions of "2016 AIME I Problems/Problem 15"
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==Solution== | ==Solution== | ||
+ | Using the radical axis theorem, the lines <math>\overline{AD}, \overline{BC}, \overline{XY}</math> are all concurrent at one point, call it <math>F</math>. Now recall by Miquel's theorem in <math>\triangle FDC</math> the fact that quadrilaterals <math>DAXY</math> and <math>CBXY</math> are cyclic implies <math>FAXB</math> is cyclic as well. Denote <math>\omega_{3}\equiv(FAXB)</math> and <math>Z\equiv\ell\cap\overline{FXY}</math>. | ||
− | + | Since point <math>Z</math> lies on the radical axis of <math>\omega_{1},\omega_{2}</math>, it has equal power with respect to both circles, thus <cmath>AZ^{2}=\text{Pow}_{\omega_{1}}(Z)=ZX\cdot ZY=\text{Pow}_{\omega_{2}}(Z)=ZB^{2}\implies AZ=ZB.</cmath> Also, notice that <cmath>AZ\cdot ZB=\text{Pow}_{\omega_{3}}(Z)=ZX\cdot ZF\implies ZY=ZF.</cmath> The diagonals of quadrilateral <math>FAYB</math> bisect each other at <math>Z</math>, so we conclude that <math>FAYB</math> is a parallelogram. Let <math>u:=ZX</math>, so that <math>ZY=ZF=u+47</math>. | |
− | |||
− | ===Solution 2=== | + | Because <math>FAYB</math> is a parallelogram and quadrilaterals <math>DAXY, CBXY</math> are cyclic, <cmath>\angle DFX=\angle AFX=\angle BYX=\angle BCX=\angle FCX~~\text{and}~~\angle XDF=\angle XDA=\angle XYA=\angle XFB=\angle XFC</cmath> so we have the pair of similar triangles <math>\triangle DFX~\sim~\triangle FCX</math>. Thus <cmath>\dfrac{37}{2u+47}=\dfrac{2u+47}{67}\implies 2u+47=\sqrt{37\cdot 67}\implies u=\dfrac{1}{2}\left(\sqrt{37\cdot 67}-47\right).</cmath> Now compute <cmath>AB^{2}=4AZ^{2}=4\cdot ZX\cdot ZY=4u(u+47)=37\cdot 67-47^{2}=\textbf{270}.</cmath> |
+ | |||
+ | [[File:AIME 2016-I15 Geogebra Diagram.png|840px]] | ||
+ | |||
+ | ==Solution 1== | ||
+ | |||
+ | Let <math>Z = XY \cap AB</math>. By the radical axis theorem <math>AD, XY, BC</math> are concurrent, say at <math>P</math>. Moreover, <math>\triangle DXP \sim \triangle PXC</math> by simple angle chasing. Let <math>y = PX, x = XZ</math>. Then <cmath>\frac{y}{37} = \frac{67}{y} \qquad \implies \qquad y^2 = 37 \cdot 67.</cmath> Now, <math>AZ^2 = \tfrac 14 AB^2</math>, and by power of a point, <cmath>\begin{align*} | ||
+ | x(y-x) &= \tfrac 14 AB^2, \quad \text{and} \\ | ||
+ | x(47+x) &= \tfrac 14 AB^2 | ||
+ | \end{align*}</cmath> Solving, we get <cmath>\tfrac 14 AB^2 = \tfrac 12 (y-47)\cdot \tfrac 12 (y+47) \qquad \implies</cmath> | ||
+ | <cmath> \qquad AB ^ 2 = 37\cdot 67 - 47^2 = \boxed{270}</cmath> | ||
+ | |||
+ | ==Solution 2== | ||
By the Radical Axis Theorem <math>AD, XY, BC</math> concur at point <math>E</math>. | By the Radical Axis Theorem <math>AD, XY, BC</math> concur at point <math>E</math>. | ||
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<cmath>XE^2 - XY^2 = (XE + XY)(XE - XY) = EY \cdot 2XS = 2SY \cdot 2SX = 4SA^2 = AB^2.</cmath>Hence <math>AB^2 = 37 \cdot 67 - 47^2 = \boxed{270}.</math> | <cmath>XE^2 - XY^2 = (XE + XY)(XE - XY) = EY \cdot 2XS = 2SY \cdot 2SX = 4SA^2 = AB^2.</cmath>Hence <math>AB^2 = 37 \cdot 67 - 47^2 = \boxed{270}.</math> | ||
− | + | ==Solution 3== | |
First, we note that as <math>\triangle XDY</math> and <math>\triangle XYC</math> have bases along the same line, <math>\frac{[\triangle XDY]}{[\triangle XYC]}=\frac{DY}{YC}</math>. We can also find the ratio of their areas using the circumradius area formula. If <math>R_1</math> is the radius of <math>\omega_1</math> and if <math>R_2</math> is the radius of <math>\omega_2</math>, then | First, we note that as <math>\triangle XDY</math> and <math>\triangle XYC</math> have bases along the same line, <math>\frac{[\triangle XDY]}{[\triangle XYC]}=\frac{DY}{YC}</math>. We can also find the ratio of their areas using the circumradius area formula. If <math>R_1</math> is the radius of <math>\omega_1</math> and if <math>R_2</math> is the radius of <math>\omega_2</math>, then | ||
<cmath>\frac{[\triangle XDY]}{[\triangle XYC]}=\frac{(37\cdot 47\cdot DY)/(4R_1)}{(47\cdot 67\cdot YC)/(4R_2)}=\frac{37\cdot DY\cdot R_2}{67\cdot YC\cdot R_1}.</cmath> | <cmath>\frac{[\triangle XDY]}{[\triangle XYC]}=\frac{(37\cdot 47\cdot DY)/(4R_1)}{(47\cdot 67\cdot YC)/(4R_2)}=\frac{37\cdot DY\cdot R_2}{67\cdot YC\cdot R_1}.</cmath> | ||
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<cmath>AB^2=30^2\cdot 37\cdot 67y^2=37\cdot 67-47^2=\boxed{270}.</cmath> | <cmath>AB^2=30^2\cdot 37\cdot 67y^2=37\cdot 67-47^2=\boxed{270}.</cmath> | ||
− | + | ==Solution 4== | |
<asy> | <asy> | ||
size(9cm); | size(9cm); | ||
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'''-Solution by TheBoomBox77''' | '''-Solution by TheBoomBox77''' | ||
+ | ==Solution 5 (not too different)== | ||
+ | Let <math>E = DA \cap CB</math>. By Radical Axes, <math>E</math> lies on <math>XY</math>. Note that <math>EAXB</math> is cyclic as <math>X</math> is the Miquel point of <math>\triangle EDC</math> in this configuration. | ||
+ | |||
+ | Claim. <math>\triangle DXE \sim \triangle EXC</math> | ||
+ | Proof. We angle chase. <cmath>\measuredangle XEC = \measuredangle XEB = \measuredangle XAB = \measuredangle XDA = \measuredangle XDE</cmath>and<cmath>\measuredangle XCE = \measuredangle XCB = \measuredangle XBA = \measuredangle XEA = \measuredangle XED. \square</cmath> | ||
+ | |||
+ | Let <math>F = EX \cap AB</math>. Note <cmath>FA^2 = FX \cdot FY = FB^2</cmath>and<cmath>EF \cdot FX = AF \cdot FB = FA^2 = FX \cdot FY \implies EF = FY</cmath>By our claim, <cmath>\frac{DX}{XE} = \frac{EX}{XC} \implies EX^2 = DX \cdot XC = 67 \cdot 37 \implies FY = \frac{EY}{2} = \frac{EX+XY}{2} = \frac{\sqrt{67 \cdot 37}+47}{2}</cmath>and<cmath>FX=FY-47=\frac{\sqrt{67 \cdot 37}-47}{2}</cmath>Finally, <cmath>AB^2 = (2 \cdot FA)^2 = 4 \cdot FX \cdot FY = 4 \cdot \frac{(67 \cdot 37) - 47^2}{4} = \boxed{270}. \blacksquare</cmath>~Mathscienceclass | ||
+ | |||
+ | |||
+ | |||
+ | ==Solution 6 (No words)== | ||
+ | [[File:2016 AIME I 15.png|600px|right]] | ||
+ | [[File:2016 AIME I 15b.png|600px|left]] | ||
+ | |||
+ | <math>AB^2 = 4 AM^2 =2x(2x+ 2 XY) =(XP - XY) (XP + XY) = XP^2 - XY^2 = XC \cdot XD - XY^2 = 67 \cdot 37 - 47^2 = \boxed{270}.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Solution 7 (Linearity of Power of a Point)== | ||
+ | Extend <math>\overline{AD}</math> and <math>\overline{BC}</math> to meet at point <math>P</math>. Let <math>M</math> be the midpoint of segment <math>AB</math>. Then by radical axis on <math>(ADY)</math>, <math>(BCY)</math> and <math>(ABCD)</math>, <math>P</math> lies on <math>XY</math>. By the bisector lemma, <math>M</math> lies on <math>XY</math>. It is well-known that <math>P</math>, <math>A</math>, <math>X</math>, and <math>B</math> are concyclic. By Power of a point on <math>M</math> with respect to <math>(PAXB)</math> and <math>(ADY)</math>, <cmath> |\text{Pow}(M, (PAXB))| = MX \cdot MP = MA^2 = |\text{Pow}(M, (ADY))| = MX \cdot MY, </cmath> so <math>MP=MY</math>. Thus <math>AB</math> and <math>PY</math> bisect each other, so <math>PAYB</math> is a parallelogram. This implies that <cmath> \angle DAY = \angle YBC, </cmath> so by the inscribed angle theorem <math>\overline{XY}</math> bisects <math>\angle DXC</math>. | ||
+ | |||
+ | Claim: <math>AB^2 = DY \cdot YC</math>. | ||
+ | |||
+ | Proof. Define the linear function <math>f(\bullet) := \text{Pow}(\bullet, (ADY)) - \text{Pow}(\bullet, (ABCD))</math>. Since <math>\overline{BY}</math> is parallel to the radical axis <math>\overline{AD}</math> of <math>(ADY)</math> and <math>(ABCD)</math> by our previous parallelism, <math>f(B)=f(Y)</math>. Note that <math>f(B)=AB^2</math> while <math>f(Y)=DY \cdot YC</math>, so we conclude. <math>\square</math> | ||
+ | |||
+ | By Stewart's theorem on <math>\triangle DXC</math>, <math>DY \cdot YC=37 \cdot 67 - 47^2 = 270</math>, so <math>AB^2=\boxed{270}</math>. | ||
+ | |||
+ | ~ Leo.Euler | ||
+ | |||
+ | ==Video Solution by MOP 2024== | ||
+ | https://youtu.be/qFfgB15fYS8 | ||
+ | |||
+ | ~r00tsOfUnity | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/QoVIorvv_I8 | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
+ | |||
+ | ==Video Solution by The Power of Logic== | ||
+ | https://youtu.be/lTZx6tp2Fvg | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2016|n=I|num-b=14|after=Last Question}} | {{AIME box|year=2016|n=I|num-b=14|after=Last Question}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 17:12, 8 January 2024
Contents
Problem
Circles and
intersect at points
and
. Line
is tangent to
and
at
and
, respectively, with line
closer to point
than to
. Circle
passes through
and
intersecting
again at
and intersecting
again at
. The three points
,
,
are collinear,
,
, and
. Find
.
Solution
Using the radical axis theorem, the lines are all concurrent at one point, call it
. Now recall by Miquel's theorem in
the fact that quadrilaterals
and
are cyclic implies
is cyclic as well. Denote
and
.
Since point lies on the radical axis of
, it has equal power with respect to both circles, thus
Also, notice that
The diagonals of quadrilateral
bisect each other at
, so we conclude that
is a parallelogram. Let
, so that
.
Because is a parallelogram and quadrilaterals
are cyclic,
so we have the pair of similar triangles
. Thus
Now compute
Solution 1
Let . By the radical axis theorem
are concurrent, say at
. Moreover,
by simple angle chasing. Let
. Then
Now,
, and by power of a point,
Solving, we get
Solution 2
By the Radical Axis Theorem concur at point
.
Let and
intersect at
. Note that because
and
are cyclic, by Miquel's Theorem
is cyclic as well. Thus
and
Thus
and
, so
is a parallelogram. Hence
and
. But notice that
and
are similar by
Similarity, so
. But
Hence
Solution 3
First, we note that as and
have bases along the same line,
. We can also find the ratio of their areas using the circumradius area formula. If
is the radius of
and if
is the radius of
, then
Since we showed this to be
, we see that
.
We extend and
to meet at point
, and we extend
and
to meet at point
as shown below.
As
is cyclic, we know that
. But then as
is tangent to
at
, we see that
. Therefore,
, and
. A similar argument shows
. These parallel lines show
. Also, we showed that
, so the ratio of similarity between
and
is
, or rather
We can now use the parallel lines to find more similar triangles. As
, we know that
Setting
, we see that
, hence
, and the problem simplifies to finding
. Setting
, we also see that
, hence
. Also, as
, we find that
As
, we see that
, hence
.
Applying Power of a Point to point with respect to
, we find
or
. We wish to find
.
Applying Stewart's Theorem to , we find
We can cancel
from both sides, finding
. Therefore,
Solution 4
First of all, since quadrilaterals
and
are cyclic, we can let
, and
, due to the properties of cyclic quadrilaterals. In addition, let
and
. Thus,
and
. Then, since quadrilateral
is cyclic as well, we have the following sums:
Cancelling out
in the second equation and isolating
yields
. Substituting
back into the first equation, we obtain
Since
we can then imply that
. Similarly,
. So then
, so since we know that
bisects
, we can solve for
and
with Stewart’s Theorem. Let
and
. Then
Now, since
and
,
. From there, let
and
. From angle chasing we can derive that
and
. From there, since
, it is quite clear that
, and
can be found similarly. From there, since
and
, we have
similarity between
,
, and
. Therefore the length of
is the geometric mean of the lengths of
and
(from
). However,
yields the proportion
; hence, the length of
is the geometric mean of the lengths of
and
.
We can now simply use arithmetic to calculate
.
-Solution by TheBoomBox77
Solution 5 (not too different)
Let . By Radical Axes,
lies on
. Note that
is cyclic as
is the Miquel point of
in this configuration.
Claim.
Proof. We angle chase.
and
Let . Note
and
By our claim,
and
Finally,
~Mathscienceclass
Solution 6 (No words)
vladimir.shelomovskii@gmail.com, vvsss
Solution 7 (Linearity of Power of a Point)
Extend and
to meet at point
. Let
be the midpoint of segment
. Then by radical axis on
,
and
,
lies on
. By the bisector lemma,
lies on
. It is well-known that
,
,
, and
are concyclic. By Power of a point on
with respect to
and
,
so
. Thus
and
bisect each other, so
is a parallelogram. This implies that
so by the inscribed angle theorem
bisects
.
Claim: .
Proof. Define the linear function . Since
is parallel to the radical axis
of
and
by our previous parallelism,
. Note that
while
, so we conclude.
By Stewart's theorem on ,
, so
.
~ Leo.Euler
Video Solution by MOP 2024
~r00tsOfUnity
Video Solution
~MathProblemSolvingSkills.com
Video Solution by The Power of Logic
See Also
2016 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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