Difference between revisions of "2022 AIME I Problems/Problem 11"
m |
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Line 5: | Line 5: | ||
defaultpen(linewidth(0.6)+fontsize(11)); | defaultpen(linewidth(0.6)+fontsize(11)); | ||
size(8cm); | size(8cm); | ||
− | pair A,B,C,D,P,Q; | + | |
+ | pair A,B,C,D,P,Q,O; | ||
A=(0,0); | A=(0,0); | ||
label("$A$", A, SW); | label("$A$", A, SW); | ||
− | B=(6 | + | B=(1.5,6*sqrt(3)); |
label("$B$", B, NW); | label("$B$", B, NW); | ||
− | C=( | + | C=(26,6*sqrt(3)); |
label("$C$", C, NE); | label("$C$", C, NE); | ||
− | D=(24,0); | + | D=(24.5,0); |
label("$D$", D, SE); | label("$D$", D, SE); | ||
− | P=( | + | P=(3.1,2.9); |
− | label("$P$", (5 | + | label("$P$", P, S); |
− | + | Q=(9.5,6); | |
− | label("$ | + | label("$Q$", Q, SE); |
+ | O=(6,3*sqrt(3)); | ||
+ | |||
+ | label("$3$", (1.3,2.2), S); | ||
+ | label("$9$", (6.5,4.3), S); | ||
+ | label("$16$", (17.5,8.9), S); | ||
+ | |||
draw(A--B--C--D--cycle); | draw(A--B--C--D--cycle); | ||
draw(C--A); | draw(C--A); | ||
− | draw( | + | draw(circle(O, 3*sqrt(3))); |
− | dot(A^^B^^C^^D | + | |
+ | dot(A^^B^^C^^D); | ||
+ | dot(intersectionpoints(circle(O, 3*sqrt(3)), A--C)); | ||
</asy> | </asy> | ||
Revision as of 00:52, 29 December 2022
Contents
[hide]Problem
Let be a parallelogram with . A circle tangent to sides , , and intersects diagonal at points and with , as shown. Suppose that , , and . Then the area of can be expressed in the form , where and are positive integers, and is not divisible by the square of any prime. Find .
Video Solution by Punxsutawney Phil
https://www.youtube.com/watch?v=1m3pqCgwLFE
Solution 1 (No trig)
Let's redraw the diagram, but extend some helpful lines.
We obviously see that we must use power of a point since they've given us lengths in a circle and there are intersection points. Let be our tangents from the circle to the parallelogram. By the secant power of a point, the power of . Then . Similarly, the power of and . We let and label the diagram accordingly.
Notice that because . Let be the center of the circle. Since and intersect and , respectively, at right angles, we have is a right-angled trapezoid and more importantly, the diameter of the circle is the height of the triangle. Therefore, we can drop an altitude from to and to , and both are equal to . Since , . Since and . We can now use Pythagorean theorem on ; we have and .
We know that because is a parallelogram. Using Pythagorean theorem on , . Therefore, base . Thus the area of the parallelogram is the base times the height, which is and the answer is
~KingRavi
Solution 2
Let the circle tangent to at separately, denote that
Using POP, it is very clear that , let , using LOC in ,, similarly, use LOC in , getting that . We use the second equation to minus the first equation, getting that , we can get .
Now applying LOC in , getting , solving this equation to get , then , , the area is leads to
~bluesoul
Solution 3
Denote by the center of the circle. Denote by the radius of the circle. Denote by , , the points that the circle meets , , at, respectively.
Because the circle is tangent to , , , , , , .
Because , , , are collinear.
Following from the power of a point, . Hence, .
Following from the power of a point, . Hence, .
Denote . Because and are tangents to the circle, .
Because is a right trapezoid, . Hence, . This can be simplified as
In , by applying the law of cosines, we have
Because , we get . Plugging this into Equation (1), we get .
Therefore,
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
Solution 4
Let be the circle, let be the radius of , and let the points at which is tangent to , , and be , , and , respectively. Note that PoP on and with respect to yields and . We can compute the area of in two ways:
1. By the half-base-height formula, .
2. We can drop altitudes from the center of to , , and , which have lengths , , and . Thus, .
Equating the two expressions for and solving for yields .
Let . By the Parallelogram Law, . Solving for yields . Thus, , for a final answer of .
~ Leo.Euler
Solution 5
Let be the circle, let be the radius of , and let the points at which is tangent to , , and be , , and , respectively. PoP on and with respect to yields
Let
In
Area is
vladimir.shelomovskii@gmail.com, vvsss
Video Solution
https://www.youtube.com/watch?v=FeM_xXiJj0c&t=1s
~Steven Chen (www.professorchenedu.com)
Video Solution 2 (Mathematical Dexterity)
https://www.youtube.com/watch?v=1nDKQkr9NaU
See Also
2022 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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.