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Difference between revisions of "2017 AMC 10A Problems/Problem 22"

(→‎Solution 2)
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(22 intermediate revisions by 7 users not shown)
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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 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===
 +
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==
 
==Solution 2==
Line 22: Line 40:
 
draw(Circle((4, 4), 4));
 
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((4-2*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("A", (4, 12.4));
 
label("B", (-.3, 6.3));
 
label("B", (-.3, 6.3));
 
label("C", (8.3, 6.3));
 
label("C", (8.3, 6.3));
 
label("O", (4, 3.4));
 
label("O", (4, 3.4));
 +
label("D", (4, 6.6));
 
</asy>
 
</asy>
 
(same diagram as Solution 1)
 
(same diagram as Solution 1)
  
−
WLOG, 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 region 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 region 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 region 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>.
+
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
 
~JH. L
 +
 +
==Video Solution by Pi Academy==
 +
 +
https://youtu.be/kWuHeHeroz0?si=xctneBvOXPE23YxC
 +
 +
~ Pi Academy
  
 
==Video Solution==
 
==Video Solution==

Latest revision as of 00:47, 1 November 2024

Problem

Sides $\overline{AB}$ and $\overline{AC}$ of equilateral triangle $ABC$ are tangent to a circle at points $B$ and $C$ respectively. What fraction of the area of $\triangle ABC$ lies outside the circle?

$\textbf{(A) } \frac{4\sqrt{3}\pi}{27}-\frac{1}{3}\qquad \textbf{(B) } \frac{\sqrt{3}}{2}-\frac{\pi}{8}\qquad \textbf{(C) } \frac{1}{2} \qquad \textbf{(D) } \sqrt{3}-\frac{2\sqrt{3}\pi}{9}\qquad \textbf{(E) } \frac{4}{3}-\frac{4\sqrt{3}\pi}{27}$

Solution 1

[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)); label("A", (4, 12.4)); label("B", (-.3, 6.3)); label("C", (8.3, 6.3)); label("O", (4, 3.4)); [/asy] Let the radius of the circle be $r$, and let its center be $O$. Since $\overline{AB}$ and $\overline{AC}$ are tangent to circle $O$, then $\angle OBA = \angle OCA = 90^{\circ}$, so $\angle BOC = 120^{\circ}$. Therefore, since $\overline{OB}$ and $\overline{OC}$ are equal to $r$, then (pick your favorite method) $\overline{BC} = r\sqrt{3}$. The area of the equilateral triangle is $\frac{(r\sqrt{3})^2 \sqrt{3}}4 = \frac{3r^2 \sqrt{3}}4$, and the area of the sector we are subtracting from it is $\frac 13 \pi r^2 - \frac 12 \cdot r\sqrt{3} \cdot \frac{r}2 = \frac{\pi r^2}3 -\frac{r^2 \sqrt{3}}4$. The area outside of the circle is $\frac{3r^2 \sqrt{3}}4-\left(\frac{\pi r^2}3 -\frac{r^2 \sqrt{3}}4\right) = r^2 \sqrt{3} - \frac{\pi r^2}3$. Therefore, the answer is \[\frac{r^2 \sqrt{3} - \frac{\pi r^2}3}{\frac{3r^2 \sqrt{3}}4} = \boxed{\textbf{(E) } \frac 43 - \frac{4\sqrt 3 \pi}{27}}\]

Note

The sector angle is $120$ because $\angle B$ and $\angle C$ are both 90 degrees meaning $\angle B + \angle C = 180^\circ$, so $ABCO$ is cyclic. Thus, the angle is $180-60=120^\circ$

~mathboy282

Alternately, $\angle ABC$ is $60^\circ$ and $\angle ABO$ is $90^\circ$, making $\angle CBO$ $30^\circ$. Symmetry allows us to use the same argument to get $\angle BCO = 30^\circ$. Since the interior angles of $\triangle BCO$ must sum to $180^\circ$, that leaves $120^\circ$ for central angle $\angle BOC$.

—wescarroll

Multiple Choice Shortcut

Assuming WLoG that the equilateral triangle's side length $s$ and therefore area $A$ are algebraic ("$\pi$-free"):

The "crust" is a circle sector minus a triangle, so its area is $a \pi - b$, where $a$ and $b$ are algebraic. Thus the answer is $(A - (a \pi -b))/A = (A+b)/A - a\pi/A$.

Once you see that $s$ is $\sqrt{3} \times$ 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 $A/(\sqrt 3)^2 = A/3$, and so the algebraic part of the answer $(A+b)/A = (A - (-A/3)) /A = 4/3$.

The transcendental ("$\pi$") part of the answer is $-a \pi /A$, and since $a$ and $A$ are algebraic, $\textbf{(E)}$ 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 $1$.

Then, the area of the triangle is $\frac{\sqrt{3}}{4}$. We are looking for the area of the portion inside the triangle but outside the circle divided by the area of the triangle. Since $m\angle ABO = 90^{\circ}=m\angle ACO = 90^{\circ}$, and $m\angle ABC = m\angle ACB = 60^{\circ}$, we know $m\angle OBC=m\angle OCB=30^{\circ}$, and $m\angle BOC = 120^{\circ}$. Drop an angle bisector of $O$ onto $BC$, call the point of intersection $D$. By SAS congruence, $\triangle BDO \cong \triangle CDO$, by CPCTC (Congruent Parts of Congruent Triangles are Congruent) $BD \cong DC$ and they both measure $\frac{1}{2}$. By 30-60-90 triangle, $OC = BO = \frac{\sqrt{3}}{3}$. The area of the sector bounded by arc BC is one-third the area of circle O, whose area is $\left(\frac{\sqrt{3}}{3}\right)^{2}\pi=\frac{1}{3}\pi$. Therefore, the area of the sector bounded by arc BC is $\frac{1}{3}\cdot\frac{1}{3}\pi=\frac{\pi}{9}$.

We are nearly there. By 30-60-90 triangle, we know $DO = \frac{\sqrt{3}}{6}$, so the area of $\triangle BOC$ is $\frac{1\cdot\frac{\sqrt{3}}{6}}{2}=\frac{\sqrt{3}}{12}$. 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 $\triangle BOC$: $\frac{\pi}{9}-\frac{\sqrt{3}}{12}$. The area of the region outside of the circle but inside the triangle is $\frac{\sqrt{3}}{4}-\left(\frac{\pi}{9}-\frac{\sqrt{3}}{12}\right)=\frac{\sqrt{3}}{4}-\frac{\pi}{9}+\frac{\sqrt{3}}{12}=\frac{\sqrt{3}}{3}-\frac{\pi}{9}$ and the ratio is $\frac{\frac{\sqrt{3}}{3}-\frac{\pi}{9}}{\frac{\sqrt{3}}{4}}=\left(\frac{\sqrt{3}}{3}-\frac{\pi}{9}\right)\cdot\frac{4}{\sqrt{3}}=\frac{4\sqrt{3}}{3\sqrt{3}}-\frac{4\pi}{9\sqrt{3}}=\frac{4}{3}-\frac{4\sqrt{3}\pi}{27} \Longrightarrow \boxed{\textbf{(E) } \frac 43 - \frac{4\sqrt 3 \pi}{27}}$.

~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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