Difference between revisions of "2017 AMC 10A Problems/Problem 22"
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<math>\textbf{(A) } \frac{4\sqrt{3}\pi}{27}-\frac{1}{3}\qquad \textbf{(B) } \frac{\sqrt{3}}{2}-\frac{\pi}{8}\qquad \textbf{(C) } \frac{1}{2} \qquad \textbf{(D) } \sqrt{3}-\frac{2\sqrt{3}\pi}{9}\qquad \textbf{(E) } \frac{4}{3}-\frac{4\sqrt{3}\pi}{27}</math> | <math>\textbf{(A) } \frac{4\sqrt{3}\pi}{27}-\frac{1}{3}\qquad \textbf{(B) } \frac{\sqrt{3}}{2}-\frac{\pi}{8}\qquad \textbf{(C) } \frac{1}{2} \qquad \textbf{(D) } \sqrt{3}-\frac{2\sqrt{3}\pi}{9}\qquad \textbf{(E) } \frac{4}{3}-\frac{4\sqrt{3}\pi}{27}</math> | ||
− | ==Solution== | + | ==Solution 1== |
<asy> | <asy> | ||
real sqrt3 = 1.73205080757; | real sqrt3 = 1.73205080757; | ||
Line 15: | Line 15: | ||
label("O", (4, 3.4)); | label("O", (4, 3.4)); | ||
</asy> | </asy> | ||
− | 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 | + | 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=== | ||
+ | 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 | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | —[[User:wescarroll|wescarroll]] | ||
+ | |||
+ | === Multiple Choice Shortcut === | ||
+ | Assuming WLoG that the equilateral triangle's side length <math>s</math> and therefore area <math>A</math> are algebraic ("<math>\pi</math>-free"): | ||
+ | |||
+ | The "crust" is a circle sector minus a triangle, so its area is <math>a \pi - b</math>, where <math>a</math> and <math>b</math> are algebraic. Thus the answer is <math>(A - (a \pi -b))/A = (A+b)/A - a\pi/A</math>. | ||
+ | |||
+ | Once you see that <math>s</math> is <math>\sqrt{3} \times</math> 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 <math>A/(\sqrt 3)^2 = A/3</math>, and so the algebraic part of the answer <math>(A+b)/A = (A - (-A/3)) /A = 4/3</math>. | ||
+ | |||
+ | The transcendental ("<math>\pi</math>") part of the answer is <math>-a \pi /A</math>, and since <math>a</math> and <math>A</math> are algebraic, <math>\textbf{(E)}</math> is the only compatible answer choice. | ||
+ | |||
+ | ==Solution 2== | ||
+ | <asy> | ||
+ | real sqrt3 = 1.73205080757; | ||
+ | draw(Circle((4, 4), 4)); | ||
+ | draw((4-2*sqrt3,6)--(4,4)--(4+2*sqrt3,6)--(4-2*sqrt3,6)--(4,12)--(4+2*sqrt3,6)); | ||
+ | draw((4, 6)--(4, 4)); | ||
+ | label("A", (4, 12.4)); | ||
+ | label("B", (-.3, 6.3)); | ||
+ | label("C", (8.3, 6.3)); | ||
+ | label("O", (4, 3.4)); | ||
+ | label("D", (4, 6.6)); | ||
+ | </asy> | ||
+ | (same diagram as Solution 1) | ||
+ | |||
+ | Without the Loss of Generality, let the side length of the triangle be <math>1</math>. | ||
+ | |||
+ | Then, the area of the triangle is <math>\frac{\sqrt{3}}{4}</math>. We are looking for the area of the portion inside the triangle but outside the circle divided by the area of the triangle. Since <math>m\angle ABO = 90^{\circ}=m\angle ACO = 90^{\circ}</math>, and <math>m\angle ABC = m\angle ACB = 60^{\circ}</math>, we know <math>m\angle OBC=m\angle OCB=30^{\circ}</math>, and <math>m\angle BOC = 120^{\circ}</math>. Drop an angle bisector of <math>O</math> onto <math>BC</math>, call the point of intersection <math>D</math>. By SAS congruence, <math>\triangle BDO \cong \triangle CDO</math>, by CPCTC (Congruent Parts of Congruent Triangles are Congruent) <math>BD \cong DC</math> and they both measure <math>\frac{1}{2}</math>. By 30-60-90 triangle, <math>OC = BO = \frac{\sqrt{3}}{3}</math>. The area of the sector bounded by arc BC is one-third the area of circle O, whose area is <math>\left(\frac{\sqrt{3}}{3}\right)^{2}\pi=\frac{1}{3}\pi</math>. Therefore, the area of the sector bounded by arc BC is <math>\frac{1}{3}\cdot\frac{1}{3}\pi=\frac{\pi}{9}</math>. | ||
+ | |||
+ | We are nearly there. By 30-60-90 triangle, we know <math>DO = \frac{\sqrt{3}}{6}</math>, so the area of <math>\triangle BOC</math> is <math>\frac{1\cdot\frac{\sqrt{3}}{6}}{2}=\frac{\sqrt{3}}{12}</math>. 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 <math>\triangle BOC</math>: <math>\frac{\pi}{9}-\frac{\sqrt{3}}{12}</math>. The area of the region outside of the circle but inside the triangle is <math>\frac{\sqrt{3}}{4}-\left(\frac{\pi}{9}-\frac{\sqrt{3}}{12}\right)=\frac{\sqrt{3}}{4}-\frac{\pi}{9}+\frac{\sqrt{3}}{12}=\frac{\sqrt{3}}{3}-\frac{\pi}{9}</math> and the ratio is <math>\frac{\frac{\sqrt{3}}{3}-\frac{\pi}{9}}{\frac{\sqrt{3}}{4}}=\left(\frac{\sqrt{3}}{3}-\frac{\pi}{9}\right)\cdot\frac{4}{\sqrt{3}}=\frac{4\sqrt{3}}{3\sqrt{3}}-\frac{4\pi}{9\sqrt{3}}=\frac{4}{3}-\frac{4\sqrt{3}\pi}{27} \Longrightarrow \boxed{\textbf{(E) } \frac 43 - \frac{4\sqrt 3 \pi}{27}}</math>. | ||
+ | |||
+ | ~JH. L | ||
==Video Solution== | ==Video Solution== |
Latest revision as of 20:45, 27 July 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
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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