Difference between revisions of "2019 Mock AMC 10B Problems/Problem 12"
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− | Observe that the triangle is a 30-60-90 triangle because of the 10 | + | We need to find the area of the intersection between the triangle and the semicircle. Observe that the triangle is a <math>30</math> - <math>60</math> - <math>90</math> triangle because of the <math>10</math> to <math>10\sqrt{3}</math> ratio. Let the center of the circle be <math>O</math>, the upper left point of the triangle be <math>A</math>, the right angle be <math>C</math>, the point on the right be <math>B</math>, and the intersection between the hypotenuse of the triangle and the curve of the semicircle be <math>I</math>. Then <math>OI</math> and <math>OB</math> are the radii of the semicircle <math>= 5\sqrt{3}</math>. Because <math>\angle ABC = 30</math> degrees, then <math>\angle BIO = 30</math> degrees (radii are the two legs of the triangle <math>\rightarrow</math> isosceles triangle). That means that <math>\angle COI</math> is <math>60</math> degrees and <math>\angle IOB</math> is <math>120</math> degrees. |
− | We know that the area of the 60-degree sector is 1 | + | To find the area of the overlapping region, we can find the area of the 60-degree sector and the area of the remaining region, which is a triangle. We know that the area of the 60-degree sector is <math>\frac{1}{6} \cdot (5\sqrt{3})^2 \pi = \frac{25}{2} \pi</math>. We also can draw the altitude of the remaining area we need to find, which forms a <math>30</math> - <math>60</math> - <math>90</math> triangle, giving us the height of the triangle is <math>\frac{15}{2}</math>. Therefore, the area of that triangular portion = <math>5\sqrt{3} \cdot \frac{15}{2} \pi \cdot \frac{1}{2} = \frac{75\sqrt{3}}{4}</math>. Therefore, the total area <math>= \frac{75\sqrt{3}}{4} + \frac{25}{2} \pi = \frac{75\sqrt{3}+50\pi}{4}</math>. Our answer <math>= 75 + 3+ 50 + 4 = \boxed{132}</math>. |
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+ | Note: <math>\gcd(a,d)=1</math> cannot be satisfied |
Latest revision as of 13:53, 11 August 2024
We need to find the area of the intersection between the triangle and the semicircle. Observe that the triangle is a -
-
triangle because of the
to
ratio. Let the center of the circle be
, the upper left point of the triangle be
, the right angle be
, the point on the right be
, and the intersection between the hypotenuse of the triangle and the curve of the semicircle be
. Then
and
are the radii of the semicircle
. Because
degrees, then
degrees (radii are the two legs of the triangle
isosceles triangle). That means that
is
degrees and
is
degrees.
To find the area of the overlapping region, we can find the area of the 60-degree sector and the area of the remaining region, which is a triangle. We know that the area of the 60-degree sector is
. We also can draw the altitude of the remaining area we need to find, which forms a
-
-
triangle, giving us the height of the triangle is
. Therefore, the area of that triangular portion =
. Therefore, the total area
. Our answer
.
Note: cannot be satisfied