Difference between revisions of "2022 AIME I Problems/Problem 11"
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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 <math>T_1, T_2, T_3</math> be our tangents from the circle to the parallelogram. By the secant power of a point, the power of <math>A = 3 \cdot (3+9) = 36</math>. Then <math>AT_2 = AT_3 = \sqrt{36} = 6</math>. Similarly, the power of <math>C = 16 \cdot (16+9) = 400</math> and <math>CT_1 = \sqrt{400} = 20</math>. We let <math>BT_3 = BT_1 = x</math> and label the diagram accordingly. | 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 <math>T_1, T_2, T_3</math> be our tangents from the circle to the parallelogram. By the secant power of a point, the power of <math>A = 3 \cdot (3+9) = 36</math>. Then <math>AT_2 = AT_3 = \sqrt{36} = 6</math>. Similarly, the power of <math>C = 16 \cdot (16+9) = 400</math> and <math>CT_1 = \sqrt{400} = 20</math>. We let <math>BT_3 = BT_1 = x</math> and label the diagram accordingly. | ||
− | Notice that because <math>BC = AD, 20+x = 6+DT_2 \implies DT_2 = 14+x</math>. Let <math>O</math> be the center of the circle. Since <math>OT_1</math> and <math>OT_2</math> intersect <math>BC</math> and <math>AD</math>, respectively, at right angles, we have <math> | + | Notice that because <math>BC = AD, 20+x = 6+DT_2 \implies DT_2 = 14+x</math>. Let <math>O</math> be the center of the circle. Since <math>OT_1</math> and <math>OT_2</math> intersect <math>BC</math> and <math>AD</math>, respectively, at right angles, we have <math>T_2T_1CD</math> 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 <math>D</math> to <math>BC</math> and <math>C</math> to <math>AD</math>, and both are equal to <math>2r</math>. Since <math>T_1E = T_2D</math>, <math>20 - CE = 14+x \implies CE = 6-x</math>. Since <math>CE = DF, DF = 6-x</math> and <math>AF = 6+14+x+6-x = 26</math>. We can now use Pythagorean theorem on <math>\triangle ACF</math>; we have <math>26^2 + (2r)^2 = (3+9+16)^2 \implies 4r^2 = 784-676 \implies 4r^2 = 108 \implies 2r = 6\sqrt{3}</math> and <math>r^2 = 27</math>. |
We know that <math>CD = 6+x</math> because <math>ABCD</math> is a parallelogram. Using Pythagorean theorem on <math>\triangle CDF</math>, <math>(6+x)^2 = (6-x)^2 + 108 \implies (6+x)^2-(6-x)^2 = 108 \implies 12 \cdot 2x = 108 \implies 2x = 9 \implies x = \frac{9}{2}</math>. Therefore, base <math>BC = 20 + \frac{9}{2} = \frac{49}{2}</math>. Thus the area of the parallelogram is the base times the height, which is <math>\frac{49}{2} \cdot 6\sqrt{3} = 147\sqrt{3}</math> and the answer is <math>\boxed{150}</math> | We know that <math>CD = 6+x</math> because <math>ABCD</math> is a parallelogram. Using Pythagorean theorem on <math>\triangle CDF</math>, <math>(6+x)^2 = (6-x)^2 + 108 \implies (6+x)^2-(6-x)^2 = 108 \implies 12 \cdot 2x = 108 \implies 2x = 9 \implies x = \frac{9}{2}</math>. Therefore, base <math>BC = 20 + \frac{9}{2} = \frac{49}{2}</math>. Thus the area of the parallelogram is the base times the height, which is <math>\frac{49}{2} \cdot 6\sqrt{3} = 147\sqrt{3}</math> and the answer is <math>\boxed{150}</math> |
Revision as of 23:14, 28 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.