Difference between revisions of "2011 AIME II Problems/Problem 10"
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− | Problem | + | == Problem 10 == |
+ | A [[circle]] with center <math>O</math> has radius 25. [[Chord]] <math>\overline{AB}</math> of length 30 and chord <math>\overline{CD}</math> of length 14 intersect at point <math>P</math>. The distance between the [[midpoint]]s of the two chords is 12. The quantity <math>OP^2</math> can be represented as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are [[relatively prime]] positive integers. Find the remainder when <math>m + n</math> is divided by 1000. | ||
− | + | == Solution 1 == | |
+ | Let <math>E</math> and <math>F</math> be the midpoints of <math>\overline{AB}</math> and <math>\overline{CD}</math>, respectively, such that <math>\overline{BE}</math> intersects <math>\overline{CF}</math>. | ||
+ | |||
+ | Since <math>E</math> and <math>F</math> are midpoints, <math>BE = 15</math> and <math>CF = 7</math>. | ||
+ | |||
+ | <math>B</math> and <math>C</math> are located on the circumference of the circle, so <math>OB = OC = 25</math>. | ||
+ | |||
+ | The line through the midpoint of a chord of a circle and the center of that circle is perpendicular to that chord, so <math>\triangle OEB</math> and <math>\triangle OFC</math> are right triangles (with <math>\angle OEB</math> and <math>\angle OFC</math> being the right angles). By the Pythagorean Theorem, <math>OE = \sqrt{25^2 - 15^2} = 20</math>, and <math>OF = \sqrt{25^2 - 7^2} = 24</math>. | ||
+ | |||
+ | Let <math>x</math>, <math>a</math>, and <math>b</math> be lengths <math>OP</math>, <math>EP</math>, and <math>FP</math>, respectively. OEP and OFP are also right triangles, so <math>x^2 = a^2 + 20^2 \to a^2 = x^2 - 400</math>, and <math>x^2 = b^2 + 24^2 \to b^2 = x^2 - 576</math> | ||
+ | |||
+ | We are given that <math>EF</math> has length 12, so, using the [[Law of Cosines]] with <math>\triangle EPF</math>: | ||
+ | |||
+ | <math>12^2 = a^2 + b^2 - 2ab \cos (\angle EPF) = a^2 + b^2 - 2ab \cos (\angle EPO + \angle FPO)</math> | ||
+ | |||
+ | Substituting for <math>a</math> and <math>b</math>, and applying the Cosine of Sum formula: | ||
+ | |||
+ | <math>144 = (x^2 - 400) + (x^2 - 576) - 2 \sqrt{x^2 - 400} \sqrt{x^2 - 576} \left( \cos \angle EPO \cos \angle FPO - \sin \angle EPO \sin \angle FPO \right)</math> | ||
+ | |||
+ | <math>\angle EPO</math> and <math>\angle FPO</math> are acute angles in right triangles, so substitute opposite/hypotenuse for sines and adjacent/hypotenuse for cosines: | ||
+ | |||
+ | <math>144 = 2x^2 - 976 - 2 \sqrt{(x^2 - 400)(x^2 - 576)} \left(\frac{\sqrt{x^2 - 400}}{x} \frac{\sqrt{x^2 - 576}}{x} - \frac{20}{x} \frac{24}{x} \right)</math> | ||
+ | |||
+ | Combine terms and multiply both sides by <math>x^2</math>: <math>144 x^2 = 2 x^4 - 976 x^2 - 2 (x^2 - 400) (x^2 - 576) + 960 \sqrt{(x^2 - 400)(x^2 - 576)}</math> | ||
+ | |||
+ | Combine terms again, and divide both sides by 64: <math>13 x^2 = 7200 - 15 \sqrt{x^4 - 976 x^2 + 230400}</math> | ||
+ | |||
+ | Square both sides: <math>169 x^4 - 187000 x^2 + 51,840,000 = 225 x^4 - 219600 x^2 + 51,840,000</math> | ||
+ | |||
+ | This reduces to <math>x^2 = \frac{4050}{7} = (OP)^2</math>; <math>4050 + 7 \equiv \boxed{057} \pmod{1000}</math>. | ||
+ | |||
+ | ==Solution 2 - Fastest== | ||
+ | |||
+ | We begin as in the first solution. Once we see that <math>\triangle EOF</math> has side lengths 12,20, and 24, we can compute its area with Heron's formula: | ||
+ | |||
+ | <math>K = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{28\cdot 16\cdot 8\cdot 4} = 32\sqrt{14}</math>. | ||
+ | |||
+ | So its circumradius is <math>R = \frac{abc}{4K} = \frac{45}{\sqrt{14}}</math>. Since <math>EPFO</math> is [[cyclic]] with diameter <math>OP</math>, we have <math>OP = 2R = \frac{90}{\sqrt{14}}</math>, so <math>OP^2 = \frac{4050}{7}</math> and the answer is <math>\boxed{057}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | We begin as the first solution have <math>OE=20</math> and <math>OF=24</math>. Because <math>\angle PEF+\angle PFO=180^{\circ}</math>, Quadrilateral <math>EPFO</math> is inscribed in a Circle. Assume point <math>I</math> is the center of this circle. | ||
+ | |||
+ | <math>\because \angle OEP=90^{\circ}</math> | ||
+ | |||
+ | <math>\therefore</math> point <math>I</math> is on <math>OP</math> | ||
+ | |||
+ | Link <math>EI</math> and <math>FI</math>, Made line <math>IK\bot EF</math>, then <math>\angle EIK=\angle EOF</math> | ||
+ | |||
+ | On the other hand, <math>\cos\angle EOF=\frac{EO^2+OF^2-EF^2}{2\cdot EO\cdot OF}=\frac{13}{15}=\cos\angle EIK</math> | ||
+ | |||
+ | <math>\sin\angle EOF=\sin\angle EIK=\sqrt{1-\frac{13^2}{15^2}}=\frac{2\sqrt{14}}{15}</math> | ||
+ | |||
+ | As a result, <math>IE=IO=\frac{45}{\sqrt 14}</math> | ||
+ | |||
+ | Therefore, <math>OP^2=4\cdot \frac{45^2}{14}=\frac{4050}{7}.</math> | ||
+ | |||
+ | As a result, <math>m+n=4057\equiv \boxed{057}\pmod{1000}</math> | ||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | Let <math>OP=x</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>. | ||
+ | |||
+ | 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} + 12\cdot x = 24\cdot \sqrt{x^2-20^2}</cmath> | ||
+ | |||
+ | Solving, we have <math>x^2=\frac{4050}{7}</math> so the answer is <math>\boxed{057}</math>. | ||
+ | |||
+ | -Solution by blueberrieejam | ||
+ | |||
+ | ~bluesoul changes the equation to a right equation, the previous equation isn't solvable | ||
+ | |||
+ | ==See also== | ||
+ | {{AIME box|year=2011|n=II|num-b=9|num-a=11}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 08:17, 21 August 2021
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 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 .
-Solution by blueberrieejam
~bluesoul changes the equation to a right equation, the previous equation isn't solvable
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.