Difference between revisions of "2016 AIME I Problems/Problem 15"
Theboombox77 (talk | contribs) (→Solution 3) |
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<cmath>37n \cdot 67n + 47^2 = 37 \cdot 67</cmath> | <cmath>37n \cdot 67n + 47^2 = 37 \cdot 67</cmath> | ||
<cmath>n^2 = \frac{270}{2479}</cmath> | <cmath>n^2 = \frac{270}{2479}</cmath> | ||
− | Now, since <math>\angle AYX = x</math> and <math>\angle BYX = y</math>, <math>\angle AYB = x + y</math>. From there, let <math>\angle AYD = \alpha</math> and <math>\angle BYC = \beta</math>. From angle chasing we can derive that <math>\angle YDX = \angle YAX = \beta - x</math> and <math>\angle YCX = \angle YBX = \alpha - y</math>. | + | Now, since <math>\angle AYX = x</math> and <math>\angle BYX = y</math>, <math>\angle AYB = x + y</math>. From there, let <math>\angle AYD = \alpha</math> and <math>\angle BYC = \beta</math>. From angle chasing we can derive that <math>\angle YDX = \angle YAX = \beta - x</math> and <math>\angle YCX = \angle YBX = \alpha - y</math>. From there, since <math>\angle ADX = x</math>, it is quite clear that <math>\angle ADY = \beta</math>, and <math>\angle YAB = \beta</math> can be found similarly. From there, since <math>\angle ADY = \angle YAB = \angle BYC = \beta</math> and <math>\angle DAY = \angle AYB = \angle YBC = x + y</math>, we have <math>AA</math> similarity between triangles <math>DAY</math>, <math>AYB</math>, and <math>YBC</math>. Therefore the length of <math>AY</math> is the geometric mean of the lengths of <math>DA</math> and <math>YB</math> (from <math>\triangle DAY \sim \triangle AYB</math>). However, <math>\triangle DAY \sim \triangle AYB \sim \triangle YBC</math> yields the proportion <math>\frac{AD}{DY} = \frac{YA}{AB} = \frac{BY}{YC}</math>; therefore, the length of <math>AB</math> is the geometric mean of the lengths of <math>DY</math> and <math>YC</math>. |
We can now simply use arithmetic to calculate <math>AB^2</math>. | We can now simply use arithmetic to calculate <math>AB^2</math>. | ||
<cmath>AB^2 = DY \cdot YC</cmath> | <cmath>AB^2 = DY \cdot YC</cmath> |
Revision as of 15:48, 16 October 2016
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
Solution 1
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 2
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 3
First of all, since quadrilaterals
and
are cyclic, we can let
, and
, due to the properties of cyclic quadrilaterals. In addition, let
and
. Then, since quadrilateral
is cyclic as well, we have the following sums:
Cancelling out
in the second equation 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 triangles
,
, and
. Therefore the length of
is the geometric mean of the lengths of
and
(from
). However,
yields the proportion
; therefore, the length of
is the geometric mean of the lengths of
and
.
We can now simply use arithmetic to calculate
.
-Solution by TheBoomBox77
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.