Difference between revisions of "2020 AMC 8 Problems/Problem 18"
(→Solution 1) |
Niuniumaths (talk | contribs) (→Video Solution by Math-X (First understand the problem!!!)) |
||
(23 intermediate revisions by 12 users not shown) | |||
Line 1: | Line 1: | ||
− | ==Problem | + | ==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> | ||
Line 10: | Line 10: | ||
<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 | + | 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 the midpoint of <math>DA</math>, so <math>OD=OA=\frac{16}{2}= 8</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>. |
− | ==Solution 2 ( | + | ==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>. | 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>. | ||
Line 26: | Line 26: | ||
draw(arc((0,0),17,360,0)); 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*SE); dot("$B$",(8,15), 1.25*N); dot("$C$",(-8,15), 1.25*N); dot("$D$",(-8,0), 1.25*SW); dot("$E$",(17,0), 1.25*E); dot("$F$",(-17,0), 1.25*W); label("$16$",(0,0),N); label("$9$",(12.5,0),N); label("$9$",(-12.5,0),N); draw((-8,-15)--(-8,0)--(8,0)--(8,-15)--cycle); dot("$B'$",(8,-15), 1.25*S); dot("$C'$",(-8,-15), 1.25*S); | draw(arc((0,0),17,360,0)); 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*SE); dot("$B$",(8,15), 1.25*N); dot("$C$",(-8,15), 1.25*N); dot("$D$",(-8,0), 1.25*SW); dot("$E$",(17,0), 1.25*E); dot("$F$",(-17,0), 1.25*W); label("$16$",(0,0),N); label("$9$",(12.5,0),N); label("$9$",(-12.5,0),N); draw((-8,-15)--(-8,0)--(8,0)--(8,-15)--cycle); dot("$B'$",(8,-15), 1.25*S); dot("$C'$",(-8,-15), 1.25*S); | ||
</asy> | </asy> | ||
− | By the [[Power of a Point Theorem]], <math>FD\cdot DE = CD\cdot C'D</math>. By symmetry, <math>CD = C'D</math>. We see that <math>FD = 9</math> and <math>DE = 25</math>. Substituting in these values, <math>9\cdot 25 = CD^2</math>, giving <math>CD^2 = 225</math> and <math>CD = 15</math>. | + | By the [[Power of a Point Theorem]], <math>FD\cdot DE = CD\cdot C'D</math>. By symmetry, <math>CD = C'D</math>. We see that <math>FD = 9</math> and <math>DE = 16 + 9 = 25</math>. Substituting in these values, <math>9\cdot 25 = CD^2</math>, giving <math>CD^2 = 225</math> and <math>CD = 15</math>. |
The area of the rectangle is therefore <math>15\cdot 16 = \boxed{\textbf{(A) }240}</math>. | The area of the rectangle is therefore <math>15\cdot 16 = \boxed{\textbf{(A) }240}</math>. | ||
+ | |||
+ | ==Solution 5 (Vertical Theorem)== | ||
+ | <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),S); label("$9$",(12.5,0),N); label("$9$",(-12.5,0),N); dot("$G$", (0,15), SE); dot("$O$", (0,0), NE); draw((0,0)--(0, 15)); draw((-7.5,15)--(0,0)); | ||
+ | </asy> | ||
+ | |||
+ | According to the Pythagorean Theorem and the Vertical Theorem, we can find out that <math> OG=\sqrt{\left(\frac{2\times9+16}{2}\right)^2 - \left(\frac{16}{2}\right)^2}=15 </math>. Therefore, the answer is <math> 15\times16=\boxed{\textbf{(A) }240} </math>; | ||
+ | |||
+ | ~[[User:Bloggish|Bloggish]] | ||
+ | |||
+ | ==Video Solution by NiuniuMaths (Easy to understand!)== | ||
+ | https://www.youtube.com/watch?v=bHNrBwwUCMI | ||
+ | |||
+ | ~NiuniuMaths | ||
+ | |||
+ | ==Video Solution by Math-X (First understand the problem!!!)== | ||
+ | https://youtu.be/UnVo6jZ3Wnk?si=_0fveg1r3WFmwmpw&t=3334 | ||
+ | |||
+ | ~Math-X | ||
+ | |||
+ | ==Video Solution (🚀 Quick 🚀)== | ||
+ | https://youtu.be/pA6Smw91Bc0 | ||
+ | |||
+ | ~Education, the Study of Everything | ||
==Video Solution by North America Math Contest Go Go Go== | ==Video Solution by North America Math Contest Go Go Go== | ||
Line 46: | Line 70: | ||
~Interstigation | ~Interstigation | ||
+ | |||
+ | ==Video Solution by OmegaLearn== | ||
+ | https://youtu.be/7SwJdAEOeAg?t=23 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution by SpreadTheMathLove== | ||
+ | https://www.youtube.com/watch?v=S_CnKCuOA_w | ||
+ | |||
+ | |||
Line 51: | Line 85: | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | >B) | ||
+ | hey ya'lll |
Latest revision as of 16:35, 26 January 2024
Contents
- 1 Problem
- 2 Solution 1 (Pythagorean Theorem)
- 3 Solution 2 (Coordinate Geometry)
- 4 Solution 3
- 5 Solution 4 (Power of a Point Theorem)
- 6 Solution 5 (Vertical Theorem)
- 7 Video Solution by NiuniuMaths (Easy to understand!)
- 8 Video Solution by Math-X (First understand the problem!!!)
- 9 Video Solution (🚀 Quick 🚀)
- 10 Video Solution by North America Math Contest Go Go Go
- 11 Video Solution by WhyMath
- 12 Video Solution
- 13 Video Solution by Interstigation
- 14 Video Solution by OmegaLearn
- 15 Video Solution by SpreadTheMathLove
Problem
Rectangle is inscribed in a semicircle with diameter
as shown in the figure. Let
and let
What is the area of
Solution 1 (Pythagorean Theorem)
Let be the center of the semicircle. The diameter of the semicircle is
, so
. By symmetry,
is 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 Sidenote: 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 formula, we scale the figure down by
; this will give the height a length of
. Now, scaling back up by
, the height
is
. The answer is then
.
-SweetMango77
Solution 4 (Power of a Point Theorem)
Draw the other half of the circle as follows:
By the Power of a Point Theorem,
. By symmetry,
. We see that
and
. Substituting in these values,
, giving
and
.
The area of the rectangle is therefore
.
Solution 5 (Vertical Theorem)
According to the Pythagorean Theorem and the Vertical Theorem, we can find out that . Therefore, the answer is
;
Video Solution by NiuniuMaths (Easy to understand!)
https://www.youtube.com/watch?v=bHNrBwwUCMI
~NiuniuMaths
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/UnVo6jZ3Wnk?si=_0fveg1r3WFmwmpw&t=3334
~Math-X
Video Solution (🚀 Quick 🚀)
~Education, the Study of Everything
Video Solution by North America Math Contest Go Go Go
https://www.youtube.com/watch?v=5Qo4pG3Uk_U
~North America Math Contest Go Go Go
Video Solution by WhyMath
~savannahsolver
Video Solution
Video Solution by Interstigation
https://youtu.be/YnwkBZTv5Fw?t=852
~Interstigation
Video Solution by OmegaLearn
https://youtu.be/7SwJdAEOeAg?t=23
~ pi_is_3.14
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=S_CnKCuOA_w
2020 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
>B) hey ya'lll