Difference between revisions of "2017 AMC 10A Problems/Problem 22"
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− | Let the radius of the circle be <math>r</math>, and let its center be <math>O</math>. Since <math>\overline{AB}</math> and <math>\overline{AC}</math> are tangent to circle <math>O</math>, then <math>\angle OBA = \angle OCA = 90^{\circ}</math>, so <math>\angle BOC = 120^{\circ}</math>. Therefore, since <math>\overline{OB}</math> and <math>\overline{OC}</math> are equal to <math>r</math>, then (pick your favorite method) <math>\overline{BC} = r\sqrt{3}</math>. The area of the equilateral triangle is <math>\frac{(r\sqrt{3})^2 \sqrt{3}}4 = \frac{3r^2 \sqrt{3}}4</math>, and the area of the sector we are subtracting from it is <math>\frac 13 \pi r^2 - \frac 12 \cdot r\sqrt{3} \cdot \frac{r}2 = \frac{\pi r^2}3 -\frac{r^2 \sqrt{3}}4</math>. The area outside of the circle is <math> \frac{3r^2 \sqrt{3}}4-\left(\frac{\pi r^2}3 -\frac{r^2 \sqrt{3}}4\right) = r^2 \sqrt{3} - \frac{\pi r^2}3</math>. Therefore, the answer is <cmath>\frac{r^2 \sqrt{3} - \frac{\pi r^2}3}{\frac{3r^2 \sqrt{3}}4} = \boxed{\textbf{(E) } \frac 43 - \frac{4\sqrt 3 \pi}{27}}</cmath> | + | Let the radius of the circle be <math>r</math>, and let its center be <math>O</math>. Since <math>\overline{AB}</math> and <math>\overline{AC}</math> are tangent to circle <math>O</math>, then <math>\angle OBA = \angle OCA = 90^{\circ}</math>, so <math>\angle BOC = 120^{\circ}</math>. Therefore, since <math>\overline{OB}</math> and <math>\overline{OC}</math> are equal to <math>r</math>, then (pick your favorite method) <math>\overline{BC} = r\sqrt{3}</math>. The area of the equilateral triangle is <math>\frac{(r\sqrt{3})^2 \sqrt{3}}4 = \frac{3r^2 \sqrt{3}}4</math>, and the area of the sector we are subtracting from it is <math>\frac 13 \pi r^2 - \frac 12 \cdot r\sqrt{3} \cdot \frac{r}2 = \frac{\pi r^2}3 -\frac{r^2 \sqrt{3}}4</math>. The area outside of the circle is <math> \frac{3r^2 \sqrt{3}}4-\left(\frac{\pi r^2}3 -\frac{r^2 \sqrt{3}}4\right) = r^2 \sqrt{3} - \frac{\pi r^2}3</math>. Therefore, the [[answer]] is <cmath>\frac{r^2 \sqrt{3} - \frac{\pi r^2}3}{\frac{3r^2 \sqrt{3}}4} = \boxed{\textbf{(E) } \frac 43 - \frac{4\sqrt 3 \pi}{27}}</cmath> |
===Note=== | ===Note=== | ||
− | The sector angle is <math>120</math> because <math>\angle B</math> and <math>\angle C</math> are both 90 degrees meaning <math>\angle B + \angle C = 180^\circ</math>, so <math>ABCO</math> is cyclic. Thus, the angle is <math>180-60=120^\circ</math> | + | The sector angle is <math>120</math> because <math>\angle B</math> and <math>\angle C</math> are both 90 degrees meaning <math>\angle B + \angle C = 180^\circ</math>, so <math>ABCO</math> is cyclic. Thus, [[the]] angle is <math>180-60=120^\circ</math> |
~mathboy282 | ~mathboy282 | ||
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+ | Alternately, <math>\angle ABC</math> is <math>60^\circ</math> and <math>\angle ABO</math> is <math>90^\circ</math>, making <math>\angle CBO</math> <math>30^\circ</math>. Symmetry allows us to use the same argument to get <math>\angle BCO = 30^\circ</math>. Since the interior angles of <math>\triangle BCO</math> must sum to <math>180^\circ</math>, that leaves <math>120^\circ</math> for [[central angle]] <math>\angle BOC</math>. | ||
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+ | —[[User:wescarroll|wescarroll]] | ||
=== Multiple Choice Shortcut === | === Multiple Choice Shortcut === | ||
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~JH. L | ~JH. L | ||
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+ | ==Video Solution by Pi Academy== | ||
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+ | https://youtu.be/kWuHeHeroz0?si=xctneBvOXPE23YxC | ||
+ | |||
+ | ~ Pi Academy | ||
==Video Solution== | ==Video Solution== |
Latest revision as of 00:47, 1 November 2024
Contents
Problem
Sides and of equilateral triangle are tangent to a circle at points and respectively. What fraction of the area of lies outside the circle?
Solution 1
Let the radius of the circle be , and let its center be . Since and are tangent to circle , then , so . Therefore, since and are equal to , then (pick your favorite method) . The area of the equilateral triangle is , and the area of the sector we are subtracting from it is . The area outside of the circle is . Therefore, the answer is
Note
The sector angle is because and are both 90 degrees meaning , so is cyclic. Thus, the angle is
~mathboy282
Alternately, is and is , making . Symmetry allows us to use the same argument to get . Since the interior angles of must sum to , that leaves for central angle .
Multiple Choice Shortcut
Assuming WLoG that the equilateral triangle's side length and therefore area are algebraic ("-free"):
The "crust" is a circle sector minus a triangle, so its area is , where and are algebraic. Thus the answer is .
Once you see that is the circle's radius, and that the circle's 30°-30°-120° triangle is two halves of an equilateral triangle, infer that the smaller circle-sector triangle's area is , and so the algebraic part of the answer .
The transcendental ("") part of the answer is , and since and are algebraic, is the only compatible answer choice.
Solution 2
(same diagram as Solution 1)
Without the Loss of Generality, let the side length of the triangle be .
Then, the area of the triangle is . We are looking for the area of the portion inside the triangle but outside the circle divided by the area of the triangle. Since , and , we know , and . Drop an angle bisector of onto , call the point of intersection . By SAS congruence, , by CPCTC (Congruent Parts of Congruent Triangles are Congruent) and they both measure . By 30-60-90 triangle, . The area of the sector bounded by arc BC is one-third the area of circle O, whose area is . Therefore, the area of the sector bounded by arc BC is .
We are nearly there. By 30-60-90 triangle, we know , so the area of is . The area of the region inside both the triangle and circle is the area of the sector bounded by arc BC minus the area of : . The area of the region outside of the circle but inside the triangle is and the ratio is .
~JH. L
Video Solution by Pi Academy
https://youtu.be/kWuHeHeroz0?si=xctneBvOXPE23YxC
~ Pi Academy
Video Solution
https://www.youtube.com/watch?v=GnJDNtjd57k&feature=youtu.be
https://youtu.be/ADDAOhNAsjQ -Video Solution by Richard Rusczyk
See Also
2017 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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 AMC 10 Problems and Solutions |
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