Difference between revisions of "2016 AIME I Problems/Problem 15"
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Circles <math>\omega_1</math> and <math>\omega_2</math> intersect at points <math>X</math> and <math>Y</math>. Line <math>\ell</math> is tangent to <math>\omega_1</math> and <math>\omega_2</math> at <math>A</math> and <math>B</math>, respectively, with line <math>AB</math> closer to point <math>X</math> than to <math>Y</math>. Circle <math>\omega</math> passes through <math>A</math> and <math>B</math> intersecting <math>\omega_1</math> again at <math>D \neq A</math> and intersecting <math>\omega_2</math> again at <math>C \neq B</math>. The three points <math>C</math>, <math>Y</math>, <math>D</math> are collinear, <math>XC = 67</math>, <math>XY = 47</math>, and <math>XD = 37</math>. Find <math>AB^2</math>. | Circles <math>\omega_1</math> and <math>\omega_2</math> intersect at points <math>X</math> and <math>Y</math>. Line <math>\ell</math> is tangent to <math>\omega_1</math> and <math>\omega_2</math> at <math>A</math> and <math>B</math>, respectively, with line <math>AB</math> closer to point <math>X</math> than to <math>Y</math>. Circle <math>\omega</math> passes through <math>A</math> and <math>B</math> intersecting <math>\omega_1</math> again at <math>D \neq A</math> and intersecting <math>\omega_2</math> again at <math>C \neq B</math>. The three points <math>C</math>, <math>Y</math>, <math>D</math> are collinear, <math>XC = 67</math>, <math>XY = 47</math>, and <math>XD = 37</math>. Find <math>AB^2</math>. | ||
− | |||
− | + | ==Solution 1== | |
Let <math>Z = XY \cap AB</math>. By the Radical Axis Theorem <math>AD, XY, BC</math> concur at point <math>E</math>. Furthermore, by simple angle chasing, <math>\triangle DXE \sim \triangle EXC</math>. Let <math>y = EX, x = XZ</math>. Then <math>\frac{y}{37} = \frac{67}{y} \implies y^2 = 37 \cdot 67</math>. Now, by Power of a Point, <math>AZ^2 = \frac{AB^2}{4}</math>, <math>(y-x)x = \frac{AB^2}{4}</math>, and <math>x(47+x) = \frac{AB^2}{4}</math>. Solving, we get <math>\dfrac{AB ^ 2}{4} = \left(\frac{y - 47}{2}\right)\left(\frac{y + 47}{2}\right) \implies AB ^ 2 = 37\cdot67 - 47^2 = \boxed{270}</math> | Let <math>Z = XY \cap AB</math>. By the Radical Axis Theorem <math>AD, XY, BC</math> concur at point <math>E</math>. Furthermore, by simple angle chasing, <math>\triangle DXE \sim \triangle EXC</math>. Let <math>y = EX, x = XZ</math>. Then <math>\frac{y}{37} = \frac{67}{y} \implies y^2 = 37 \cdot 67</math>. Now, by Power of a Point, <math>AZ^2 = \frac{AB^2}{4}</math>, <math>(y-x)x = \frac{AB^2}{4}</math>, and <math>x(47+x) = \frac{AB^2}{4}</math>. Solving, we get <math>\dfrac{AB ^ 2}{4} = \left(\frac{y - 47}{2}\right)\left(\frac{y + 47}{2}\right) \implies AB ^ 2 = 37\cdot67 - 47^2 = \boxed{270}</math> | ||
− | + | ==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. | 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. | ||
Revision as of 01:42, 9 December 2021
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 1
Let . By the Radical Axis Theorem
concur at point
. Furthermore, by simple angle chasing,
. Let
. Then
. Now, by Power of a Point,
,
, and
. 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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.