Difference between revisions of "2016 AIME I Problems/Problem 15"
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− | 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 = | + | 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 = XY</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(y-x) &= \tfrac 14 AB^2, \quad \text{and} \\ | ||
x(47+x) &= \tfrac 14 AB^2 | x(47+x) &= \tfrac 14 AB^2 |
Revision as of 01:33, 29 July 2022
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
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)
Shelomovskii, vvsss, www.deoma-cmd.ru
Video Solution
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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