Difference between revisions of "2020 AMC 8 Problems/Problem 18"
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+ | ==Problem== | ||
Rectangle <math>ABCD</math> is inscribed in a semicircle with diameter <math>\overline{FE},</math> as shown in the figure. Let <math>DA=16,</math> and let <math>FD=AE=9.</math> What is the area of <math>ABCD?</math> | Rectangle <math>ABCD</math> is inscribed in a semicircle with diameter <math>\overline{FE},</math> as shown in the figure. Let <math>DA=16,</math> and let <math>FD=AE=9.</math> What is the area of <math>ABCD?</math> | ||
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==Solution 1== | ==Solution 1== | ||
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<asy> draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot("$A$",(8,0), 1.25*S); dot("$B$",(8,15), 1.25*N); dot("$C$",(-8,15), 1.25*N); dot("$D$",(-8,0), 1.25*S); dot("$E$",(17,0), 1.25*S); dot("$F$",(-17,0), 1.25*S); label("$16$",(0,0),N); label("$9$",(12.5,0),N); label("$9$",(-12.5,0),N); dot("$O$", (0,0), 1.25*S); draw((0,0)--(-8,15));</asy> | <asy> draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot("$A$",(8,0), 1.25*S); dot("$B$",(8,15), 1.25*N); dot("$C$",(-8,15), 1.25*N); dot("$D$",(-8,0), 1.25*S); dot("$E$",(17,0), 1.25*S); dot("$F$",(-17,0), 1.25*S); label("$16$",(0,0),N); label("$9$",(12.5,0),N); label("$9$",(-12.5,0),N); dot("$O$", (0,0), 1.25*S); draw((0,0)--(-8,15));</asy> | ||
− | + | Let <math>O</math> be the center of the semicircle. The diameter of the semicircle is <math>9+16+9=34</math>, so <math>OC = 17</math>. By symmetry, <math>O</math> is in fact the midpoint of <math>DA</math>, so <math>OD=OA=\frac{16}{2}= </math>. By the Pythagorean theorem in right-angled triangle <math>ODC</math> (or <math>OBA</math>), we have that <math>CD</math> (or <math>AB</math>) is <math>\sqrt{17^2-8^2}=15</math>. Accordingly, the area of <math>ABCD</math> is <math>16\cdot 15=\boxed{\textbf{(A) }240}</math>. | |
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− | ==Solution | + | ==Solution 2 (coordinate geometry)== |
+ | Let the midpoint of segment <math>FE</math> be the origin. Evidently, point <math>D=(-8,0)</math> and <math>A=(8,0)</math>. Since points <math>C</math> and <math>B</math> share <math>x</math>-coordinates with <math>D</math> and <math>A</math> respectively, it suffices to find the <math>y</math>-coordinate of <math>B</math> (which will be the height of the rectangle) and multiply this by <math>DA</math> (which we know is <math>16</math>). The radius of the semicircle is <math>\frac{9+16+9}{2} = 17</math>, so the whole circle has equation <math>x^2+y^2=289</math>; as already stated, <math>B</math> has the same <math>x</math>-coordinate as <math>A</math>, i.e. <math>8</math>, so substituting this into the equation shows that <math>y=\pm15</math>. Since <math>y>0</math> at <math>B</math>, the y-coordinate of <math>B</math> is <math>15</math>. Therefore, the answer is <math>16\cdot 15 = \boxed{\textbf{(A) }240}</math>. | ||
− | + | (Note that the synthetic solution (Solution 1) is definitely faster and more elegant. However, this is the solution that you should use if you can't see any other easier strategy.) | |
− | + | ==Solution 3== | |
+ | We can use a result from the Art of Problem Solving <i>Introduction to Algebra</i> book: for a semicircle with diameter <math>(1+n)</math>, such that the <math>1</math> part is on one side and the <math>n</math> part is on the other side, the height from the end of the <math>1</math> side (or the start of the <math>n</math> side) is <math>\sqrt{n}</math>. To use this, we scale the figure down by <math>9</math>; then the height is <math>\sqrt{1+\frac{16}{9}} = \sqrt{\frac{16+9}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3}</math>. Now, scaling back up by <math>9</math>, the height <math>DC</math> is <math>9 \cdot \frac{5}{3} = 15</math>. The answer is now <math>15 \cdot 16 = \boxed{\textbf{(A) }240}</math>. | ||
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{{AMC8 box|year=2020|num-b=17|num-a=19}} | {{AMC8 box|year=2020|num-b=17|num-a=19}} | ||
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[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 09:41, 20 November 2020
Problem
Rectangle is inscribed in a semicircle with diameter as shown in the figure. Let and let What is the area of
Solution 1
Let be the center of the semicircle. The diameter of the semicircle is , so . By symmetry, is in fact the midpoint of , so . By the Pythagorean theorem in right-angled triangle (or ), we have that (or ) is . Accordingly, the area of is .
Solution 2 (coordinate geometry)
Let the midpoint of segment be the origin. Evidently, point and . Since points and share -coordinates with and respectively, it suffices to find the -coordinate of (which will be the height of the rectangle) and multiply this by (which we know is ). The radius of the semicircle is , so the whole circle has equation ; as already stated, has the same -coordinate as , i.e. , so substituting this into the equation shows that . Since at , the y-coordinate of is . Therefore, the answer is .
(Note that the synthetic solution (Solution 1) is definitely faster and more elegant. However, this is the solution that you should use if you can't see any other easier strategy.)
Solution 3
We can use a result from the Art of Problem Solving Introduction to Algebra book: for a semicircle with diameter , such that the part is on one side and the part is on the other side, the height from the end of the side (or the start of the side) is . To use this, we scale the figure down by ; then the height is . Now, scaling back up by , the height is . The answer is now .
2020 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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