Difference between revisions of "2021 AMC 12A Problems/Problem 24"
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==Diagram== | ==Diagram== | ||
− | [asy | + | [[File: 2021 AMC 12A Problem 24.png|center]] |
+ | ~MRENTHUSIASM (by Geometry Expressions) | ||
+ | |||
+ | ==Solution 1 (Possible Without Trigonometry)== | ||
+ | Let <math>O=\Gamma</math> be the center of the semicircle and <math>X=\Omega</math> be the center of the circle. | ||
+ | |||
+ | Applying the Extended Law of Sines to <math>\triangle PQR,</math> we find the radius of <math>\odot X:</math> <cmath>XP=\frac{QR}{2\cdot\sin \angle QPR}=\frac{3\sqrt3}{2\cdot\frac{\sqrt3}{2}}=3.</cmath> | ||
+ | Alternatively, by the Inscribed Angle Theorem, <math>\triangle QRX</math> is a <math>30^\circ\text{-}30^\circ\text{-}120^\circ</math> triangle with base <math>QR=3\sqrt3.</math> Dividing <math>\triangle QRX</math> into two congruent <math>30^\circ\text{-}60^\circ\text{-}90^\circ</math> triangles, we get that the radius of <math>\odot X</math> is <math>XQ=XR=3</math> by the side-length ratios. | ||
+ | |||
+ | Let <math>M</math> be the midpoint of <math>\overline{QR}.</math> By the Perpendicular Chord Bisector Converse, we have <math>\overline{XM}\perp\overline{QR}</math> and <math>\overline{OM}\perp\overline{QR}.</math> Together, points <math>O, X,</math> and <math>M</math> must be collinear. | ||
+ | |||
+ | By the SAS Congruence, we have <math>\triangle QXM\cong\triangle RXM,</math> both of which are <math>30^\circ\text{-}60^\circ\text{-}90^\circ</math> triangles. By the side-length ratios, <math>RM=\frac{3\sqrt3}{2}, RX=3,</math> and <math>XM=\frac{3}{2}.</math> By the Pythagorean Theorem on right <math>\triangle ORM,</math> we get <math>OM=\frac{13}{2}</math> and <math>OX=OM-XM=5.</math> By the Pythagorean Theorem on right <math>\triangle OXP,</math> we obtain <math>OP=4.</math> | ||
+ | [[File: 2021 AMC 12A Problem 24(2) (Revised).png|center]] | ||
+ | As shown above, we construct altitude <math>\overline{PC}</math> of <math>\triangle PQR.</math> Since <math>\overline{PC}\perp\overline{QR}</math> and <math>\overline{OM}\perp\overline{QR},</math> we know that <math>\overline{PC}\parallel\overline{OM}.</math> We construct point <math>D</math> on <math>\overline{PC}</math> such that <math>\overline{XD}\perp\overline{PC}.</math> Clearly, quadrilateral <math>MXDC</math> is a rectangle. Since <math>\angle XPD=\angle OXP</math> by alternate interior angles, we have <math>\triangle XPD\sim\triangle OXP</math> by the AA Similarity, with the ratio of similitude <math>\frac{XP}{OX}=\frac 35.</math> Therefore, we get <math>PD=\frac 95</math> and <math>PC=PD+DC=PD+XM=\frac 95 + \frac 32 = \frac{33}{10}.</math> | ||
+ | |||
+ | The area of <math>\triangle PQR</math> is <cmath>\frac12\cdot QR\cdot PC=\frac12\cdot3\sqrt3\cdot\frac{33}{10}=\frac{99\sqrt3}{20},</cmath> from which the answer is <math>99+3+20=\boxed{\textbf{(D) } 122}.</math> | ||
+ | |||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | ==Solution 2 (Trigonometry)== | ||
+ | <asy> | ||
draw(circle((7,0),7)); | draw(circle((7,0),7)); | ||
pair A = (0, 0); | pair A = (0, 0); | ||
Line 11: | Line 31: | ||
draw(A--B); | draw(A--B); | ||
draw(circle((11,3),3)); | draw(circle((11,3),3)); | ||
− | label(" | + | label("$C$", (7, 0), S); |
− | label(" | + | label("$O$", (11, 3), E); |
− | label(" | + | label("$P$", (11, 0), S); |
pair C = (7, 0); | pair C = (7, 0); | ||
pair O = (11, 3); | pair O = (11, 3); | ||
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draw(P--Q); | draw(P--Q); | ||
draw(P--R); | draw(P--R); | ||
− | label(" | + | label("$Q$", Q, N); |
− | label(" | + | label("$R$", R, E); |
− | + | </asy> | |
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− | |||
Suppose we label the points as shown in the diagram above, where <math>C</math> is the center of the semicircle and <math>O</math> is the center of the circle tangent to <math>\overline{AB}</math>. Since <math>\angle QPR = 60^{\circ}</math>, we have <math>\angle QOR = 2\cdot 60^{\circ}=120^{\circ}</math> and <math>\triangle QOR</math> is a <math>30-30-120</math> triangle, which can be split into two <math>30-60-90</math> triangles by the altitude from <math>O</math>. Since <math>QR=3\sqrt{3},</math> we know <math>OQ=OR=\tfrac{3\sqrt{3}}{\sqrt{3}}=3</math> by <math>30-60-90</math> triangles. The area of this part of <math>\triangle PQR</math> is <math>\frac{1}{2}bh=\tfrac{3\sqrt{3}}{2}\cdot\tfrac{3}{2}=\tfrac{9\sqrt{3}}{4}</math>. We would like to add this value to the sum of the areas of the other two parts of <math>\triangle PQR</math>. | Suppose we label the points as shown in the diagram above, where <math>C</math> is the center of the semicircle and <math>O</math> is the center of the circle tangent to <math>\overline{AB}</math>. Since <math>\angle QPR = 60^{\circ}</math>, we have <math>\angle QOR = 2\cdot 60^{\circ}=120^{\circ}</math> and <math>\triangle QOR</math> is a <math>30-30-120</math> triangle, which can be split into two <math>30-60-90</math> triangles by the altitude from <math>O</math>. Since <math>QR=3\sqrt{3},</math> we know <math>OQ=OR=\tfrac{3\sqrt{3}}{\sqrt{3}}=3</math> by <math>30-60-90</math> triangles. The area of this part of <math>\triangle PQR</math> is <math>\frac{1}{2}bh=\tfrac{3\sqrt{3}}{2}\cdot\tfrac{3}{2}=\tfrac{9\sqrt{3}}{4}</math>. We would like to add this value to the sum of 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>. | 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 | |
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− | ~ | ||
− | === | + | ==Solution 3 (Weighted Averages and Similar Triangles)== |
− | + | <asy> unitsize(25); pair C = (7, 0); draw(arc(C, 7, 0, 180)); pair A = (0, 0), B = (14, 0); draw(A--B); draw(circle((11,3),3)); label("$A$", A, SSE); label("$B$", B, SSW); label("$C$", (A+B)/2, S); label("$O$", (11, 3), E); label("$P$", (11, 0), S); pair O = (11, 3), P = (11, 0), Q = intersectionpoints(circle(C, 7), circle(O, 3))[1], R = intersectionpoints(circle(C, 7), circle(O, 3))[0], S = (Q+R)/2, N = (121/8, 0), T = (8/11)*N + (3/11)*R, X = (4/7)*T + (3/7)*S; draw(C--O, blue); draw(O--S, red); draw(C--Q); draw(C--R); draw(Q--N--B); draw(O--P); draw(O--Q); draw(O--R); draw(P--Q--R--cycle); draw(B--T); draw(P--X); label("$Q$", Q, NNE); label("$R$", R, E); label("$S$", S, ENE); label("$N$", N, SSE); label("$T$", T, ENE); label("$X$", X, dir(20)); draw(rightanglemark(P, X, Q)); draw(rightanglemark(B, T, R)); draw(rightanglemark(C, S, Q)); </asy> | |
+ | Define points as shown above, where <math>N=\overleftrightarrow{PA}\cap\overleftrightarrow{QR}</math>. The area of <math>\triangle PQR</math> is simply<cmath>\dfrac{1}{2}PX\cdot QR=\dfrac{3\sqrt{3}}{2}PX;</cmath> | ||
+ | it remains to compute the value of <math>PX</math>. Note that <math>PX</math> is simply a weighted average of <math>BT</math> and <math>CS;</math> it is <math>\dfrac{CP}{BP}</math> times closer to <math>BT</math> than it is to <math>CS</math>. Observe that<cmath>CS=\sqrt{CQ^{2}-\left(\dfrac{1}{2}QR\right)^{2}}=\sqrt{7^{2}-\left(\dfrac{3\sqrt{3}}{2}\right)^{2}}=6.5</cmath> | ||
+ | since the radius of <math>\Gamma</math> is <math>7</math> as its diameter is <math>14</math>. Note also by the Extended Law of Sines the radius of <math>\Omega</math> is <math>\dfrac{3\sqrt{3}}{2\sin 60^{\circ}}=3,</math> so <math>OS=3\cos 60^{\circ}=1.5</math>. Since <math>C, O,</math> and <math>S</math> are collinear by symmetry we have <math>CO=CS-OS=5,</math> so <math>CP=\sqrt{5^{2}-3^{2}}=4</math> and <math>BP=7-4=3</math>. Therefore, <math>\triangle OPC</math> is a <math>3\text{-}4\text{-}5</math> right triangle; <math>\triangle OPC\sim\triangle NSC</math> since <math>\angle OPC=\angle CSN=90^{\circ}</math> and <math>\angle OCP=\angle NCS=\sin^{-1}\left(\dfrac{3}{5}\right)</math>. Therefore <math>\dfrac{CN}{CS}=\cfrac{CO}{CP}=\dfrac{5}{4}</math> so <math>CN=\dfrac{5}{4}CS=\dfrac{65}{8}</math>. Since <math>\triangle BTN\sim\triangle CSN,</math> we have <math>\dfrac{BT}{BN}=\dfrac{CS}{CN}=\dfrac{4}{5}</math>. Therefore <cmath>BT=\dfrac{4}{5}BN=\dfrac{4}{5}\left(CN-7\right)=\dfrac{4}{5}\cdot\dfrac{9}{8}=\dfrac{36}{40}=0.9;</cmath> | ||
+ | so <math>PX</math> is <math>\dfrac{4}{3}</math> times as close to <math>0.9</math> as to <math>6.5;</math> we can compute <math>PX=\dfrac{4}{7}BT+\dfrac{3}{7}CS=\dfrac{4}{7}\cdot0.9+\dfrac{3}{7}\cdot6.5=3.3</math>. The area of <math>\triangle PQR</math> is<cmath>\dfrac{3\sqrt{3}}{2}\cdot 3.3=\dfrac{99\sqrt{3}}{20}</cmath>and <math>99+3+20=\boxed{\textbf{(D)} ~122}</math>. | ||
− | + | ~sugar_rush | |
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− | ~ | ||
==Video Solution by Punxsutawney Phil== | ==Video Solution by Punxsutawney Phil== | ||
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== Video Solution by OmegaLearn (Similar Triangles, Law of Sines, Law of Cosines ) == | == Video Solution by OmegaLearn (Similar Triangles, Law of Sines, Law of Cosines ) == | ||
https://youtu.be/j965v6ahUZk | https://youtu.be/j965v6ahUZk | ||
+ | |||
+ | ~pi_is_3.14 | ||
==See also== | ==See also== |
Latest revision as of 17:57, 17 May 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 ?
Diagram
~MRENTHUSIASM (by Geometry Expressions)
Solution 1 (Possible Without Trigonometry)
Let be the center of the semicircle and be the center of the circle.
Applying the Extended Law of Sines to we find the radius of Alternatively, by the Inscribed Angle Theorem, is a triangle with base Dividing into two congruent triangles, we get that the radius of is by the side-length ratios.
Let be the midpoint of By the Perpendicular Chord Bisector Converse, we have and Together, points and must be collinear.
By the SAS Congruence, we have both of which are triangles. By the side-length ratios, and By the Pythagorean Theorem on right we get and By the Pythagorean Theorem on right we obtain
As shown above, we construct altitude of Since and we know that We construct point on such that Clearly, quadrilateral is a rectangle. Since by alternate interior angles, we have by the AA Similarity, with the ratio of similitude Therefore, we get and
The area of is from which the answer is
~MRENTHUSIASM
Solution 2 (Trigonometry)
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
Solution 3 (Weighted Averages and Similar Triangles)
Define points as shown above, where . The area of is simply it remains to compute the value of . Note that is simply a weighted average of and it is times closer to than it is to . Observe that since the radius of is as its diameter is . Note also by the Extended Law of Sines the radius of is so . Since and are collinear by symmetry we have so and . Therefore, is a right triangle; since and . Therefore so . Since we have . Therefore so is times as close to as to we can compute . The area of isand .
~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 )
~pi_is_3.14
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.