Difference between revisions of "2011 AIME II Problems/Problem 10"
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This reduces to <math>x^2 = \frac{4050}{7} = (OP)^2</math>; <math>4050 + 7 \equiv \boxed{057} \pmod{1000}</math>. | This reduces to <math>x^2 = \frac{4050}{7} = (OP)^2</math>; <math>4050 + 7 \equiv \boxed{057} \pmod{1000}</math>. | ||
− | ==Solution 2== | + | ==Solution 2 - Fastest== |
− | We begin as in the first solution. Once we see that <math>\triangle | + | 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. | |
+ | |||
+ | 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 | ||
+ | |||
+ | ==Solution 5 (Quick Angle Solution)== | ||
+ | Let <math>M</math> be the midpoint of <math>AB</math> and <math>N</math> of <math>CD</math>. As <math>\angle OMP = \angle ONP</math>, quadrilateral <math>OMPN</math> is cyclic with diameter <math>OP</math>. By Cyclic quadrilaterals note that <math>\angle MPO = \angle MNO</math>. | ||
+ | |||
+ | The area of <math>\triangle MNP</math> can be computed by Herons as <cmath>[MNO] = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{28\cdot 16\cdot 8\cdot 4} = 32\sqrt{14}.</cmath> The area is also <math>\frac{1}{2}ON \cdot MN \sin{\angle MNO}</math>. Therefore, | ||
+ | <cmath>\begin{align*} | ||
+ | \sin{\angle MNO} &= \frac{2[MNO]}{ON \cdot MN} \ | ||
+ | &= \frac{2}{9}\sqrt{14} \ | ||
+ | \sin{\angle MNO} &= \frac{OM}{OP} \ | ||
+ | &= \frac{2}{9}\sqrt{14} \ | ||
+ | OP &= \frac{90\sqrt{14}}{14} \ | ||
+ | OP^2 &= \frac{4050}{7} \implies \boxed{057}. | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | ~ 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 contains no quadratic equations (unless you count LoC). | ||
+ | |||
+ | ==Solution 7 Analytic Geometry== | ||
+ | [[Image:2011 AIMEII Problem 10 CASE 2.png|525px]] | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | Since <math>\overline{OE}\perp \overline{AB}</math> and <math>\overline{OF}\perp \overline{CD}</math>, | ||
+ | <math>OE = \sqrt{OB^2-BE^2}=20</math> and <math>OF = \sqrt{OC^2-OF^2}=24</math> | ||
+ | |||
+ | With law of cosines, <math>\cos \angle EOF = \frac{OE^2+OF^2-EF^2}{2\cdot OE\cdot OF} = \frac{13}{15}</math> | ||
+ | |||
+ | Since <math>EF < OF</math>, <math>\angle EOF</math> is acute angle. <math>\sin \angle EOF = \sqrt{1-\cos^2 \angle EOF} = \frac{\sqrt{56}}{15}</math> and <math>\tan \angle EOF = \frac{\sqrt{56}}{13}</math> | ||
+ | |||
+ | Let <math>\overline{OF}</math> line be <math>x</math> axis. | ||
+ | |||
+ | Line <math>\overline{DC}</math> equation is <math>x = OF</math>. | ||
+ | |||
+ | Since line <math>\overline{AB}</math> passes point <math>E</math> and perpendicular to <math>\overline{OD}</math>, its equation is <math>y - E_y = -\frac{1}{\tan \angle EOF} (x - E_x)</math> | ||
+ | |||
+ | where <math>E_x = OE\cos{\angle EOF}</math> , <math>E_y = OE\sin{\angle EOF}</math> | ||
+ | |||
+ | Since <math>P</math> is the intersection of <math>\overline{AB}</math> and <math>\overline{CD}</math>, | ||
+ | |||
+ | <math>P_x = OF = 24</math> | ||
+ | |||
+ | <math>P_y = E_y -\frac{1}{\tan \angle EOF} (OF - E_x) = - \frac{3\sqrt{14}}{7}</math> (Negative means point <math>P</math> is between point <math>F</math> and <math>C</math>) | ||
+ | |||
+ | <math>OP^2 = P_x^2 + P_y^2 = \frac{4050}{7}</math> and the answer is <math>\boxed{057}</math>. | ||
+ | |||
+ | Note: if <math>EF</math> was longer, point <math>P</math> would be between point <math>D</math> and <math>F</math>. Then, <math>OP</math> would be the diagonal of quadrilateral <math>OEPF</math> not the side. To apply Ptolemy's theorem like solution 4, it is critical to know whether <math>OP</math> is the diagonal or side of quadrilateral. Equation for wrong case cannot be solved. For example, | ||
+ | |||
+ | [[Image:2011 AIMEII Problem 10 CASE 1.png|525px]] | ||
==See also== | ==See also== |
Revision as of 02:12, 18 December 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
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 contains no quadratic equations (unless you count LoC).
Solution 7 Analytic Geometry
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 .
Since and , and
With law of cosines,
Since , is acute angle. and
Let line be axis.
Line equation is .
Since line passes point and perpendicular to , its equation is
where ,
Since is the intersection of and ,
(Negative means point is between point and )
and the answer is .
Note: if was longer, point would be between point and . Then, would be the diagonal of quadrilateral not the side. To apply Ptolemy's theorem like solution 4, it is critical to know whether is the diagonal or side of quadrilateral. Equation for wrong case cannot be solved. For example,
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.