Difference between revisions of "2011 AIME II Problems/Problem 10"
Mr.sharkman (talk | contribs) |
Treehead123 (talk | contribs) (Added an analytic geometry solution with diagrams) |
||
Line 122: | Line 122: | ||
-Mr.Sharkman | -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 diagonal. For example, | ||
+ | |||
+ | [[Image:2011 AIMEII Problem 10 CASE 1.png|525px]] | ||
− | |||
==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 01:23, 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 diagonal. 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.