Difference between revisions of "2011 AIME II Problems/Problem 10"
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~ Aaryabhatta1 | ~ Aaryabhatta1 | ||
+ | |||
+ | ==Solution 6== | ||
+ | Define <math>M</math> and <math>N</math> as the midpoints of <math>AB</math> and <math>CD</math>, respectively. Because <math>\angle OMP = \angle ONP = 90^{\circ}</math>, we have that <math>ONPM</math> is a cyclic quadrilateral. Hence, <math>\angle PNM = \angle POM.</math> Then, let these two angles be denoted as <math>\alpha</math>. | ||
+ | Now, assume WLOG that <math>PD = x < 7</math> and <math>PB = y < 15</math> (We can do this because one of <math>PD</math> or <math>PC</math> must be less than 7, and similarly for <math>PB</math> and <math>PA</math>). Then, by Power of a Point on P with respect to the circle with center <math>O</math>, we have that | ||
+ | <cmath>(14-x)x = (30-y)y</cmath> | ||
+ | <cmath>(7-x)^{2}+176=(15-y)^{2}.</cmath> | ||
+ | Then, let <math>z = (7-x)^{2}</math>. From Law of Cosines on <math>\triangle NMP</math>, we have that | ||
+ | <cmath>\textrm{cos } \angle MNP = \frac{NP^{2}+MN^{2}-MP^{2}}{2 \cdot NP \cdot MN} </cmath> | ||
+ | <cmath>\textrm{cos } \alpha = \frac{(7-x)^{2} + 12^{2} - (14-x)^{2}}{24 \cdot (7-x)}.</cmath> | ||
+ | Plugging in <math>z</math> in gives | ||
+ | <cmath>\textrm{cos } \alpha = \frac{-32}{24 \cdot \sqrt{z}}</cmath> | ||
+ | <cmath>\textrm{cos } \alpha = \frac{-4}{3\sqrt{z}}</cmath> | ||
+ | <cmath>\textrm{cos }^{2} \alpha = \frac{16}{9z}.</cmath> | ||
+ | Hence, | ||
+ | <cmath>\textrm{tan }^{2} \alpha = \frac{\frac{9z-16}{9z}}{\frac{16}{9z}} = \frac{9z-16}{16}.</cmath> | ||
+ | Then, we also know that | ||
+ | <cmath>\textrm{tan } \alpha = \textrm{tan } \angle MOP = \frac{MP}{OM} = \frac{14-y}{20}.</cmath> | ||
+ | Squaring this, we get | ||
+ | <cmath>\textrm{tan }^{2} \alpha = \frac{z+176}{400}.</cmath> | ||
+ | Equating our expressions for <math>z</math>, we get | ||
+ | <math>\frac{z+176}{400} = \frac{9z-16}{16}.</math> | ||
+ | Solving gives us that | ||
+ | <math>z = \frac{18}{7}</math>. | ||
+ | Since <math>\angle ONP = 90^{\circ}</math>, from the Pythagorean Theorem, | ||
+ | <math>OP^{2} = ON^{2}+PN^{2} = 25^{2}-7^{2} + z = 576+z = \frac{4050}{7}</math>, | ||
+ | and thus the answer is <math>4050+7 = 4057</math>, which when divided by a thousand leaves a remainder of <math>\boxed{57}.</math> | ||
+ | |||
+ | -Mr.Sharkman | ||
+ | |||
+ | |||
+ | Note: my solution was very long and tedious. It was definitely was the least elegant solution. The only thing I like about it is it's the only solution without a quadratic equation. | ||
==See also== | ==See also== | ||
{{AIME box|year=2011|n=II|num-b=9|num-a=11}} | {{AIME box|year=2011|n=II|num-b=9|num-a=11}} |
Revision as of 07:13, 11 August 2023
Contents
Problem 10
A circle with center has radius 25. Chord
of length 30 and chord
of length 14 intersect at point
. The distance between the midpoints of the two chords is 12. The quantity
can be represented as
, where
and
are relatively prime positive integers. Find the remainder when
is divided by 1000.
Solution 1
Let and
be the midpoints of
and
, respectively, such that
intersects
.
Since and
are midpoints,
and
.
and
are located on the circumference of the circle, so
.
The line through the midpoint of a chord of a circle and the center of that circle is perpendicular to that chord, so and
are right triangles (with
and
being the right angles). By the Pythagorean Theorem,
, and
.
Let ,
, and
be lengths
,
, and
, respectively. OEP and OFP are also right triangles, so
, and
We are given that has length 12, so, using the Law of Cosines with
:
Substituting for and
, and applying the Cosine of Sum formula:
and
are acute angles in right triangles, so substitute opposite/hypotenuse for sines and adjacent/hypotenuse for cosines:
Combine terms and multiply both sides by :
Combine terms again, and divide both sides by 64:
Square both sides:
This reduces to ;
.
Solution 2 - Fastest
We begin as in the first solution. Once we see that has side lengths
,
, and
, we can compute its area with Heron's formula:
Thus, the circumradius of triangle is
. Looking at
, we see that
, which makes it a cyclic quadrilateral. This means
's circumcircle and
's inscribed circle are the same.
Since is cyclic with diameter
, we have
, so
and the answer is
.
Solution 3
We begin as the first solution have and
. Because
, Quadrilateral
is inscribed in a Circle. Assume point
is the center of this circle.
point
is on
Link and
, Made line
, then
On the other hand,
As a result,
Therefore,
As a result,
Solution 4
Let .
Proceed as the first solution in finding that quadrilateral has side lengths
,
,
, and
, and diagonals
and
.
We note that quadrilateral is cyclic and use Ptolemy's theorem to solve for
:
Solving, we have so the answer is
.
-Solution by blueberrieejam
~bluesoul changes the equation to a right equation, the previous equation isn't solvable
Solution 5 (Quick Angle Solution)
Let be the midpoint of
and
of
. As
, quadrilateral
is cyclic with diameter
. By Cyclic quadrilaterals note that
.
The area of can be computed by Herons as
The area is also
. Therefore,
~ Aaryabhatta1
Solution 6
Define and
as the midpoints of
and
, respectively. Because
, we have that
is a cyclic quadrilateral. Hence,
Then, let these two angles be denoted as
.
Now, assume WLOG that
and
(We can do this because one of
or
must be less than 7, and similarly for
and
). Then, by Power of a Point on P with respect to the circle with center
, we have that
Then, let
. From Law of Cosines on
, we have that
Plugging in
in gives
Hence,
Then, we also know that
Squaring this, we get
Equating our expressions for
, we get
Solving gives us that
.
Since
, from the Pythagorean Theorem,
,
and thus the answer is
, which when divided by a thousand leaves a remainder of
-Mr.Sharkman
Note: my solution was very long and tedious. It was definitely was the least elegant solution. The only thing I like about it is it's the only solution without a quadratic equation.
See also
2011 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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