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Difference between revisions of "2019 Mock AMC 10B Problems/Problem 12"
 
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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 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.
 
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>.
 
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 12:53, 11 August 2024

We need to find the area of the intersection between the triangle and the semicircle. Observe that the triangle is a $30$ - $60$ - $90$ triangle because of the $10$ to $10\sqrt{3}$ ratio. Let the center of the circle be $O$, the upper left point of the triangle be $A$, the right angle be $C$, the point on the right be $B$, and the intersection between the hypotenuse of the triangle and the curve of the semicircle be $I$. Then $OI$ and $OB$ are the radii of the semicircle $= 5\sqrt{3}$. Because $\angle ABC = 30$ degrees, then $\angle BIO = 30$ degrees (radii are the two legs of the triangle $\rightarrow$ isosceles triangle). That means that $\angle COI$ is $60$ degrees and $\angle IOB$ is $120$ 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 $\frac{1}{6} \cdot (5\sqrt{3})^2 \pi = \frac{25}{2} \pi$. We also can draw the altitude of the remaining area we need to find, which forms a $30$ - $60$ - $90$ triangle, giving us the height of the triangle is $\frac{15}{2}$. Therefore, the area of that triangular portion = $5\sqrt{3} \cdot \frac{15}{2} \pi \cdot \frac{1}{2} = \frac{75\sqrt{3}}{4}$. Therefore, the total area $= \frac{75\sqrt{3}}{4} + \frac{25}{2} \pi = \frac{75\sqrt{3}+50\pi}{4}$. Our answer $= 75 + 3+ 50 + 4 = \boxed{132}$.

Note: $\gcd(a,d)=1$ cannot be satisfied

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