Difference between revisions of "2011 AIME II Problems/Problem 10"
(added solution 4) |
(→Solution 4) |
||
Line 64: | Line 64: | ||
Let <math>OP=x</math>. | Let <math>OP=x</math>. | ||
− | Proceed as the first solution in finding that quadrilateral <math> | + | |
− | We note that quadrilateral <math> | + | Proceed as the first solution in finding that quadrilateral <math>EPFO</math> has side lengths <math>OE=20</math>, <math>OF=24</math>, <math>EP=\sqrt{x^2-20^2}</math>, and <math>PF=\sqrt{x^2-24^2}</math>, and diagonals <math>OP=x</math> and <math>EF=12</math>. |
− | < | + | |
− | Solving | + | We note that quadrilateral <math>EPFO</math> is cyclic and use Ptolemy's theorem to solve for <math>x</math>: |
+ | |||
+ | <cmath>20\cdot \sqrt{x^2-24^2} + 24\cdot \sqrt{x^2-20^2} = 12\cdot x</cmath> | ||
+ | |||
+ | Solving, we have <math>x^2=\frac{4050}{7}</math> so the answer is <math>\boxed{057}</math>. | ||
==See also== | ==See also== |
Revision as of 12:35, 29 May 2020
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
We begin as in the first solution. Once we see that has side lengths 12,20, and 24, we can compute its area with Heron's formula:
.
So its circumradius is . 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
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.