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=== |
Revision as of 19:39, 27 July 2024
Contents
[hide]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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.