Difference between revisions of "2021 AMC 12B Problems/Problem 24"
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<math>\textbf{(A) }81 \qquad \textbf{(B) }89 \qquad \textbf{(C) }97\qquad \textbf{(D) }105 \qquad \textbf{(E) }113</math> | <math>\textbf{(A) }81 \qquad \textbf{(B) }89 \qquad \textbf{(C) }97\qquad \textbf{(D) }105 \qquad \textbf{(E) }113</math> | ||
− | == | + | ==Solution 1== |
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Let <math>X</math> denote the intersection point of the diagonals <math>AC</math> and <math>BD</math>. Remark that by symmetry <math>X</math> is the midpoint of both <math>\overline{PQ}</math> and <math>\overline{RS}</math>, so <math>XP = XQ = 3</math> and <math>XR = XS = 4</math>. Now note that since <math>\angle APB = \angle ARB = 90^\circ</math>, quadrilateral <math>ARPB</math> is cyclic, and so | Let <math>X</math> denote the intersection point of the diagonals <math>AC</math> and <math>BD</math>. Remark that by symmetry <math>X</math> is the midpoint of both <math>\overline{PQ}</math> and <math>\overline{RS}</math>, so <math>XP = XQ = 3</math> and <math>XR = XS = 4</math>. Now note that since <math>\angle APB = \angle ARB = 90^\circ</math>, quadrilateral <math>ARPB</math> is cyclic, and so | ||
<cmath>XR\cdot XA = XP\cdot XB,</cmath>which implies <math>\tfrac{XA}{XB} = \tfrac{XP}{XR} = \tfrac34</math>. | <cmath>XR\cdot XA = XP\cdot XB,</cmath>which implies <math>\tfrac{XA}{XB} = \tfrac{XP}{XR} = \tfrac34</math>. | ||
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Thus let <math>x> 0</math> be such that <math>XA = 3x</math> and <math>XB = 4x</math>. Then Pythagorean Theorem on <math>\triangle APX</math> yields <math>AP = \sqrt{AX^2 - XP^2} = 3\sqrt{x^2-1}</math>, and so<cmath>[ABCD] = 2[ABD] = AP\cdot BD = 3\sqrt{x^2-1}\cdot 8x = 24x\sqrt{x^2-1}.</cmath>Solving this for <math>x^2</math> yields <math>x^2 = \tfrac12 + \tfrac{\sqrt{41}}8</math>, and so<cmath>(8x)^2 = 64x^2 = 64\left(\tfrac12 + \tfrac{\sqrt{41}}8\right) = 32 + 8\sqrt{41}.</cmath>The requested answer is <math>32 + 8 + 41 = \boxed{81}</math>. | Thus let <math>x> 0</math> be such that <math>XA = 3x</math> and <math>XB = 4x</math>. Then Pythagorean Theorem on <math>\triangle APX</math> yields <math>AP = \sqrt{AX^2 - XP^2} = 3\sqrt{x^2-1}</math>, and so<cmath>[ABCD] = 2[ABD] = AP\cdot BD = 3\sqrt{x^2-1}\cdot 8x = 24x\sqrt{x^2-1}.</cmath>Solving this for <math>x^2</math> yields <math>x^2 = \tfrac12 + \tfrac{\sqrt{41}}8</math>, and so<cmath>(8x)^2 = 64x^2 = 64\left(\tfrac12 + \tfrac{\sqrt{41}}8\right) = 32 + 8\sqrt{41}.</cmath>The requested answer is <math>32 + 8 + 41 = \boxed{81}</math>. | ||
− | ==Solution 2(Trig) == | + | ==Solution 2 (Trig) == |
− | Let <math>X</math> denote the intersection point of the diagonals <math>AC</math> and <math>BD</math> | + | Let <math>X</math> denote the intersection point of the diagonals <math>AC</math> and <math>BD,</math> and let <math>\theta = \angle{COB}</math>. Then, by the given conditions, <math>XR = 4,</math> <math>XQ = 3,</math> <math>[XCB] = \frac{15}{4}</math>. So, |
<cmath> XC = \frac{3}{\cos \theta}</cmath> | <cmath> XC = \frac{3}{\cos \theta}</cmath> | ||
<cmath> XB \cos \theta = 4 </cmath> | <cmath> XB \cos \theta = 4 </cmath> | ||
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Combining the above 3 equations, we get | Combining the above 3 equations, we get | ||
<cmath>\frac{\sin \theta }{\cos^2 \theta} = \frac{5}{8}.</cmath> | <cmath>\frac{\sin \theta }{\cos^2 \theta} = \frac{5}{8}.</cmath> | ||
− | Since we want to find <math>d^2 = 4XB^2 = \frac{64}{\cos^2 \theta}</math> | + | Since we want to find <math>d^2 = 4XB^2 = \frac{64}{\cos^2 \theta},</math> we let <math>x = \frac{1}{\cos^2 \theta}.</math> Then |
<cmath> \frac{\sin^2 \theta }{\cos^4 \theta} = \frac{1-\cos ^2 \theta}{\cos^4 \theta} = x^2 - x = \frac{25}{64}.</cmath> | <cmath> \frac{\sin^2 \theta }{\cos^4 \theta} = \frac{1-\cos ^2 \theta}{\cos^4 \theta} = x^2 - x = \frac{25}{64}.</cmath> | ||
− | Solving this, we get <math>x = \frac{4 + \sqrt{41}}{8}</math> | + | Solving this, we get <math>x = \frac{4 + \sqrt{41}}{8},</math> so <math>d^2 = 64x = 32 + 8\sqrt{41}</math>. <math>\boxed{81}</math> |
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+ | ==Solution 3 (Similar Triangles and Algebra)== | ||
+ | Let <math>X</math> be the intersection of diagonals <math>AC</math> and <math>BD</math>. By symmetry <math>[\triangle XCB] = \frac{15}{4}</math>, <math>XQ = 3</math> and <math>XR = 4</math>, so now we have reduced all of the conditions one quadrant. Let <math>CQ = x</math>. <math>XC = \sqrt{x^2+9}</math>, <math>RB = \frac{4x}{3}</math> by similar triangles and using the area condition we get <math>\frac{4}{3} \cdot x \cdot \sqrt{x^2+9} = \frac{15}{2}</math>. Note that it suffices to find <math>OB = \frac{4}{3}\sqrt{x^2+9}</math> because we can double and square it to get <math>d^2</math>. Solving for <math>a = x^2</math> in the above equation, and then using <math>d^2 = \frac{64}{9}(x^2+9) = 8\sqrt{41} + 32 \Rightarrow \boxed{81}</math>. | ||
− | + | ==Solution 4 (Similar Triangles)== | |
+ | Again, Let <math>X</math> be the intersection of diagonals <math>AC</math> and <math>BD</math>. Note that triangles <math>\triangle QXC</math> and <math>\triangle BXR</math> are similar because they are right triangles and share <math>\angle CXQ</math>. First, call the length of <math>XB = \frac{d}{2}</math>. By the definition of an area of a parallelogram, <math>CQ \cdot 2XB = 15</math>, so <math>CQ = \frac{15}{d}</math>. Using similar triangles on <math>\triangle QXC</math> and <math>\triangle BXR</math>, <math>\frac{CQ}{XQ} = \frac{BR}{XR}</math>. Therefore, finding <math>BR</math>, <math>BR = \frac{XR}{XQ} \cdot CQ = \frac{4}{3} \cdot \frac{15}{d} = \frac{20}{d}</math>. Now, applying the Pythagorean theorem once, we find <math>(\frac{20}{d}) ^2</math> + <math>(4)^2</math> = <math>(\frac{d}{2}) ^2</math>. Solving this equation for <math>d^2</math>, we find <math>d^2=\frac{64+\sqrt{4096+6400}}{2}=32+8\sqrt{41} \Rightarrow \boxed{81}</math>. | ||
== Video Solution by OmegaLearn (Cyclic Quadrilateral and Power of a Point) == | == Video Solution by OmegaLearn (Cyclic Quadrilateral and Power of a Point) == | ||
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~ pi_is_3.14 | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution (Simple: Using trigonometry and Equations)== | ||
+ | https://youtu.be/ZB-VN02H6mU | ||
+ | ~hippopotamus1 | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2021|ab=B|num-b=23|num-a=25}} | {{AMC12 box|year=2021|ab=B|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 19:10, 17 May 2021
Contents
Problem
Let be a parallelogram with area . Points and are the projections of and respectively, onto the line and points and are the projections of and respectively, onto the line See the figure, which also shows the relative locations of these points.
Suppose and and let denote the length of the longer diagonal of Then can be written in the form where and are positive integers and is not divisible by the square of any prime. What is
Solution 1
Let denote the intersection point of the diagonals and . Remark that by symmetry is the midpoint of both and , so and . Now note that since , quadrilateral is cyclic, and so which implies .
Thus let be such that and . Then Pythagorean Theorem on yields , and soSolving this for yields , and soThe requested answer is .
Solution 2 (Trig)
Let denote the intersection point of the diagonals and and let . Then, by the given conditions, . So, Combining the above 3 equations, we get Since we want to find we let Then Solving this, we get so .
Solution 3 (Similar Triangles and Algebra)
Let be the intersection of diagonals and . By symmetry , and , so now we have reduced all of the conditions one quadrant. Let . , by similar triangles and using the area condition we get . Note that it suffices to find because we can double and square it to get . Solving for in the above equation, and then using .
Solution 4 (Similar Triangles)
Again, Let be the intersection of diagonals and . Note that triangles and are similar because they are right triangles and share . First, call the length of . By the definition of an area of a parallelogram, , so . Using similar triangles on and , . Therefore, finding , . Now, applying the Pythagorean theorem once, we find + = . Solving this equation for , we find .
Video Solution by OmegaLearn (Cyclic Quadrilateral and Power of a Point)
~ pi_is_3.14
Video Solution (Simple: Using trigonometry and Equations)
https://youtu.be/ZB-VN02H6mU ~hippopotamus1
See Also
2021 AMC 12B (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.