Difference between revisions of "2021 AMC 12A Problems/Problem 24"
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To find the areas of the other two parts of <math>\triangle PQR</math> using the <math>\sin</math> area formula, we need the sides and included angles. Here we know the sides but what we don't know are the angles. So it seems like we will have to use an angle from another triangle and combine them with the angles we already know to find these angles easily. We know that <math>\angle QOR = 120^{\circ}</math> and triangles <math>\triangle COQ</math> and <math>\triangle COR</math> are congruent as they share a side, <math>CQ=CR,</math> and <math>OQ=OR</math>. Therefore <math>\angle COQ = \angle COR = 120^{\circ}</math>. Suppose <math>CO=x</math>. Then <math>3^{2}+x^{2}-6x\cos{120^{\circ}}=7^{2}</math>, and since <math>\cos{120^{\circ}}=-\tfrac{1}{2}</math>, this simplifies to <math>x^{2}+3x=7^{2}-3^{2}\rightarrow x^{2}+3x-40=0</math>. This factors nicely as <math>(x-5)(x+8)=0</math>, so <math>x=5</math> as <math>x</math> can't be <math>-8</math>. Since <math>CO=5, OP=3</math> and <math>\angle OPC=90^{\circ}</math>, we now know that <math>\triangle OPC</math> is a <math>3-4-5</math> right triangle. This may be useful info for later as we might use an angle in this triangle to find the areas of the other two parts of <math>\triangle PQR</math>. | To find the areas of the other two parts of <math>\triangle PQR</math> using the <math>\sin</math> area formula, we need the sides and included angles. Here we know the sides but what we don't know are the angles. So it seems like we will have to use an angle from another triangle and combine them with the angles we already know to find these angles easily. We know that <math>\angle QOR = 120^{\circ}</math> and triangles <math>\triangle COQ</math> and <math>\triangle COR</math> are congruent as they share a side, <math>CQ=CR,</math> and <math>OQ=OR</math>. Therefore <math>\angle COQ = \angle COR = 120^{\circ}</math>. Suppose <math>CO=x</math>. Then <math>3^{2}+x^{2}-6x\cos{120^{\circ}}=7^{2}</math>, and since <math>\cos{120^{\circ}}=-\tfrac{1}{2}</math>, this simplifies to <math>x^{2}+3x=7^{2}-3^{2}\rightarrow x^{2}+3x-40=0</math>. This factors nicely as <math>(x-5)(x+8)=0</math>, so <math>x=5</math> as <math>x</math> can't be <math>-8</math>. Since <math>CO=5, OP=3</math> and <math>\angle OPC=90^{\circ}</math>, we now know that <math>\triangle OPC</math> is a <math>3-4-5</math> right triangle. This may be useful info for later as we might use an angle in this triangle to find the areas of the other two parts of <math>\triangle PQR</math>. | ||
− | Let <math>\angle POC = \alpha</math>. Then <math>\sin\alpha = \tfrac{4}{5}, \cos\alpha = \tfrac{3}{5}, \angle QOP = 120+\alpha,</math> and <math>\angle POR = 120-\alpha</math>. The sum of the areas of <math>\triangle QOP</math> and <math>\triangle POR</math> is <math>3\cdot 3\cdot\tfrac{1}{2}\cdot\left[\sin(120-\alpha)+\sin(120+\alpha)\right]=\tfrac{9}{2}\left[\sin(120-\alpha)+\sin(120+\alpha)\right],</math> which we will add to <math>\tfrac{9\sqrt{3}}{4}</math> to get the area of <math>\triangle PQR</math>. Observe that <cmath>\sin(120-\alpha) = \sin 120\cos\alpha-\sin\alpha\cos 120 = \tfrac{\sqrt{3}}{2}\cdot\tfrac{3}{5}-\tfrac{4}{5}\cdot\tfrac{-1}{2}=\tfrac{3\sqrt{3}}{10}+\tfrac{4}{10}=\tfrac{3\sqrt{3}+4}{10}</cmath>and similarly <math>\sin(120+\alpha)=\tfrac{3\sqrt{3}-4}{10}</math>. Adding these two gives <math>\tfrac{3\sqrt{3}}{5}</math> and multiplying that by <math>\tfrac{9}{2}</math> gets us <math>\tfrac{27\sqrt{3}}{10},</math> which we add to <math>\tfrac{9\sqrt{3}}{4}</math> to get <math>\tfrac{54\sqrt{3}+45\sqrt{3}}{20}=\tfrac{99\sqrt{3}}{20}</math>. The answer is <math>99+3+20=102+20=\boxed{\textbf{(D)} ~122}</math> | + | Let <math>\angle POC = \alpha</math>. Then <math>\sin\alpha = \tfrac{4}{5}, \cos\alpha = \tfrac{3}{5}, \angle QOP = 120+\alpha,</math> and <math>\angle POR = 120-\alpha</math>. The sum of the areas of <math>\triangle QOP</math> and <math>\triangle POR</math> is <math>3\cdot 3\cdot\tfrac{1}{2}\cdot\left[\sin(120-\alpha)+\sin(120+\alpha)\right]=\tfrac{9}{2}\left[\sin(120-\alpha)+\sin(120+\alpha)\right],</math> which we will add to <math>\tfrac{9\sqrt{3}}{4}</math> to get the area of <math>\triangle PQR</math>. Observe that <cmath>\sin(120-\alpha) = \sin 120\cos\alpha-\sin\alpha\cos 120 = \tfrac{\sqrt{3}}{2}\cdot\tfrac{3}{5}-\tfrac{4}{5}\cdot\tfrac{-1}{2}=\tfrac{3\sqrt{3}}{10}+\tfrac{4}{10}=\tfrac{3\sqrt{3}+4}{10}</cmath>and similarly <math>\sin(120+\alpha)=\tfrac{3\sqrt{3}-4}{10}</math>. Adding these two gives <math>\tfrac{3\sqrt{3}}{5}</math> and multiplying that by <math>\tfrac{9}{2}</math> gets us <math>\tfrac{27\sqrt{3}}{10},</math> which we add to <math>\tfrac{9\sqrt{3}}{4}</math> to get <math>\tfrac{54\sqrt{3}+45\sqrt{3}}{20}=\tfrac{99\sqrt{3}}{20}</math>. The answer is <math>99+3+20=102+20=\boxed{\textbf{(D)} ~122}.</math> |
+ | |||
+ | ~sugar_rush | ||
==Video Solution by Punxsutawney Phil== | ==Video Solution by Punxsutawney Phil== |
Revision as of 20:25, 23 February 2021
Contents
Problem
Semicircle has diameter of length . Circle lies tangent to at a point and intersects at points and . If and , then the area of equals , where and are relatively prime positive integers, and is a positive integer not divisible by the square of any prime. What is ?
Solution 1
Diagram
~MRENTHUSIASM (by Geometry Expressions)
Solution
Let be the center of the semicircle, be the center of the circle, and be the midpoint of By the Perpendicular Chord Theorem Converse, we have and Together, points and must be collinear.
Applying the Extended Law of Sines on we havein which the radius of is
By the SAS Congruence, we have both of which are -- triangles. By the side-length ratios, and By the Pythagorean Theorem in we get and By the Pythagorean Theorem on we obtain
As shown above, we construct an altitude of Since and we know that We construct on such that Clearly, is a rectangle. Since by alternate interior angles, we have by the AA Similarity, with ratio of similitude Therefore, we get that and
The area of isand the answer is
~MRENTHUSIASM
Solution 2
Diagram
[asy] draw(circle((7,0),7)); pair A = (0, 0); pair B = (14, 0); draw(A--B); draw(circle((11,3),3)); label("", (7, 0), S); label("", (11, 3), E); label("", (11, 0), S); pair C = (7, 0); pair O = (11, 3); pair P = (11, 0); pair Q = intersectionpoints(circle(C, 7), circle(O, 3))[1]; pair R = intersectionpoints(circle(C, 7), circle(O, 3))[0]; draw(C--O); draw(C--Q); draw(C--R); draw(Q--R); draw(O--P); draw(O--Q); draw(O--R); draw(P--Q); draw(P--R); label("", Q, N); label("", R, E); [/asy]
someone pls make this asymptote code work
Solution
Suppose we label the points as shown in the diagram above, where is the center of the semicircle and is the center of the circle tangent to . Since , we have and is a triangle, which can be split into two triangles by the altitude from . Since we know by triangles. The area of this part of is . We would like to add this value to the sum of the areas of the other two parts of .
To find the areas of the other two parts of using the area formula, we need the sides and included angles. Here we know the sides but what we don't know are the angles. So it seems like we will have to use an angle from another triangle and combine them with the angles we already know to find these angles easily. We know that and triangles and are congruent as they share a side, and . Therefore . Suppose . Then , and since , this simplifies to . This factors nicely as , so as can't be . Since and , we now know that is a right triangle. This may be useful info for later as we might use an angle in this triangle to find the areas of the other two parts of .
Let . Then and . The sum of the areas of and is which we will add to to get the area of . Observe that and similarly . Adding these two gives and multiplying that by gets us which we add to to get . The answer is
~sugar_rush
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=cEHF5iWMe9c
Video Solution by OmegaLearn (Similar Triangles, Law of Sines, Law of Cosines )
See also
2021 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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