Difference between revisions of "2011 AIME II Problems/Problem 10"
Gladiasked (talk | contribs) m (→Solution 2 - Fastest) |
Gladiasked (talk | contribs) m (→Solution 2 - Fastest) |
||
Line 37: | Line 37: | ||
We begin as in the first solution. Once we see that <math>\triangle EOF</math> has side lengths <math>12</math>, <math>20</math>, and <math>24</math>, we can compute its area with Heron's formula: | We begin as in the first solution. Once we see that <math>\triangle EOF</math> has side lengths <math>12</math>, <math>20</math>, and <math>24</math>, we can compute its area with Heron's formula: | ||
− | < | + | <cmath>K = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{28\cdot 16\cdot 8\cdot 4} = 32\sqrt{14}.</cmath> |
Thus, the circumradius of triangle <math>\triangle EOF</math> is <math>R = \frac{abc}{4K} = \frac{45}{\sqrt{14}}</math>. Looking at <math>EPFO</math>, we see that <math>\angle OEP = \angle OFP = 90^\circ</math>, which makes it a cyclic quadrilateral. This means <math>\triangle EOF</math>'s circumcircle and <math>EPFO</math>'s inscribed circle are the same. | Thus, the circumradius of triangle <math>\triangle EOF</math> is <math>R = \frac{abc}{4K} = \frac{45}{\sqrt{14}}</math>. Looking at <math>EPFO</math>, we see that <math>\angle OEP = \angle OFP = 90^\circ</math>, which makes it a cyclic quadrilateral. This means <math>\triangle EOF</math>'s circumcircle and <math>EPFO</math>'s inscribed circle are the same. |
Revision as of 11:37, 3 June 2023
Contents
[hide]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
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.