Difference between revisions of "2017 AMC 10A Problems/Problem 22"
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=== Multiple Choice Shortcut === | === Multiple Choice Shortcut === | ||
+ | Assuming WLoG that the equilateral triangle's side length <math>s</math> and therefore area <math>A</math> are algebraic ("\pi<math>-free"): | ||
− | Once you see that | + | 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>. |
− | The transcendental ("<math>\pi< | + | |
+ | 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$ are algebraic, \textbf{(E)} is the only compatible answer choice. | ||
==Solution 2== | ==Solution 2== |
Revision as of 08:20, 5 April 2023
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
Multiple Choice Shortcut
Assuming WLoG that the equilateral triangle's side length and therefore area
are algebraic ("\pi$-free"):
The "crust" is a circle sector minus a triangle, so its area is$ (Error compiling LaTeX. Unknown error_msg)a \pi - ba
b
(A - (a \pi -b))/A = (A+b)/A - a\pi/A$.
Once you see that$ (Error compiling LaTeX. Unknown error_msg)s\sqrt{3} \times
A/(\sqrt 3)^2 = A/3
(A+b)/A = (A - (-A/3)) /A = 4/3$.
The transcendental ("$ (Error compiling LaTeX. Unknown error_msg)\pi-a \pi /A
a
A$ are algebraic, \textbf{(E)} 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.