Difference between revisions of "2020 AMC 8 Problems/Problem 18"
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− | Let the midpoint of segment <math> | + | Let the midpoint of segment <math>EF</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.) | (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.) |
Revision as of 18:45, 2 January 2023
Contents
Problem 18
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 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 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
.
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
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 |
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