Difference between revisions of "2024 AMC 10A Problems/Problem 14"
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draw((12*sqrt(3),4)--(12+12*sqrt(3),4)); | draw((12*sqrt(3),4)--(12+12*sqrt(3),4)); | ||
label("$12$",(6+12*sqrt(3),4),1.5S); | label("$12$",(6+12*sqrt(3),4),1.5S); | ||
+ | dot(A^^B^^C,linewidth(4)); | ||
</asy> | </asy> | ||
+ | |||
+ | ~MRENTHUSIASM | ||
== Solution 1 == | == Solution 1 == | ||
− | |||
− | |||
− | |||
− | |||
− | <math>a+b+c</math> is <math>48+3+24</math> which is <math>\textbf{(D)}~75</math> | + | Call the bottom vertices <math>B</math> and <math>C</math> (the one closer to the circle is <math>C</math>) and the top vertex <math>A</math>. The tangency point between the circle and the side of the triangle is <math>D</math>, and the tangency point on line <math>\ell</math> <math>E</math>, and the center of the circle is <math>O</math> |
+ | |||
+ | |||
+ | Draw radii to the tangency points, the arc is <math>60</math> degrees because <math>\angle ACB</math> is <math>60</math>, and since <math>\angle DCE</math> is supplementary, it's <math>120^{\circ}</math>. The sum of the angles in a quadrilateral is <math>360</math>, which means <math>\angle COD</math> is <math>60^{\circ}</math> | ||
+ | |||
+ | |||
+ | Triangle ODC is <math>30</math>-<math>60</math>-<math>90</math> triangle so CD is <math>4\sqrt{3}</math>. | ||
+ | Since we have <math>2</math> congruent triangles (<math>\triangle ODC</math> and <math>\triangle OEC</math>), the combined area of both is <math>48\sqrt{3}</math>. | ||
+ | The area of the arc is <math>144 \cdot \frac{60}{360} \cdot \pi</math> which is <math>24\pi</math>, so the answer is <math>48\sqrt{3}-24\pi</math> | ||
+ | |||
+ | |||
+ | <math>a+b+c</math> is <math>48+3+24</math> which is <math>\boxed{\textbf{(D)}~75}</math> | ||
~ASPALAPATI75 | ~ASPALAPATI75 | ||
+ | ~andy_liu766 (latex) | ||
+ | |||
+ | edits by KR | ||
+ | |||
+ | ==Note== | ||
+ | There were two possible configurations from this problem; the one described in the solution above and the configuration in which the circle is tangent to the bottom of line <math>\ell</math> and the base of the equilateral triangle. However, since the area in this configuration is simply <math>0,</math> we can infer that the problem is talking about the configuration in Solution 1. | ||
+ | |||
+ | ~dbnl | ||
+ | |||
+ | ==Solution 2 (Quick Guess)== | ||
+ | |||
+ | Since this problem involves equilateral triangles, the only possible number under the square root is <math>3</math>. Now subtracting all of the answer choices by <math>3</math>, we get: | ||
+ | <cmath>\textbf{(A)}~72-3=69\qquad\textbf{(B)}~73-3=70\qquad\textbf{(C)}~74-3=71\qquad\textbf{(D)}~75-3=72\qquad\textbf{(E)}~76-3=73</cmath> | ||
+ | |||
+ | Due to the even parity of the problem, we can safely assume that the answer is either <math>B</math> or <math>D</math>, but as <math>D</math> is a multiple of <math>12</math> and <math>24</math>, we get the answer of <math>\boxed{\textbf{(D)}~75}</math>. | ||
+ | |||
+ | ~megaboy6679 | ||
+ | |||
+ | ==Solution 3== | ||
+ | (pardon the diagrams :D) | ||
+ | |||
+ | say the area we want to find is x. | ||
+ | |||
+ | since the equilateral triangle has an internal angle of 60, the exterior angle formed by the triangle and the line is 120. simplifying the diagram you will get: | ||
+ | |||
+ | \ ##### | ||
+ | \ ######## | ||
+ | \######## | ||
+ | \#####___________ | ||
+ | |||
+ | make three of these that each circle is tangent to the other 2 circles | ||
+ | |||
+ | \ | ||
+ | \ ##### | ||
+ | \ ######## | ||
+ | #####\######## | ||
+ | #######\_####__________ | ||
+ | #######/ ##### | ||
+ | #####/######## | ||
+ | / ######## | ||
+ | / ###### | ||
+ | |||
+ | Since they are 3 congruent triangles, you can make an equilateral triangle using their radius(12), with each vertex at the center of each circle. This will make an equilateral triangle of side length 24. if you look now, the area within the equilateral triangle consists of 3 <math>60/360</math> of a circle, and 3 of x. | ||
+ | |||
+ | we first find the area of the triangle, which is <math>24 * 12\sqrt{3}</math>, we then find the area of <math>60/360</math> of a circle, which is <math>60/360 * 12^2\pi</math>, we subtract <math>24 * 12\sqrt{3}</math> by <math>60/360 * 12^2\pi</math>, and divide by 3, yielding the area of x. | ||
+ | |||
+ | <math>24 * 12\sqrt{3}</math> - <math>60/360 * 12^2\pi</math> | ||
+ | _________________________________________ = <math>48\sqrt{3}-24\pi</math> | ||
+ | 3 | ||
+ | |||
+ | |||
+ | <math>a+b+c</math> is <math>48+3+24</math> which is <math>\boxed{\textbf{(D)}~75}</math> | ||
+ | |||
+ | |||
+ | ~Yiguo Zhang | ||
+ | |||
+ | == Video Solution by Pi Academy == | ||
+ | |||
+ | https://youtu.be/ABkKz0gS1MU?si=ZQBgDMRaJmMPSSMM | ||
+ | |||
− | + | == Video Solution 1 by Power Solve == | |
+ | https://youtu.be/oCQ_QvXqV5s | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2024|ab=A|num-b=13|num-a=15}} | {{AMC10 box|year=2024|ab=A|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 14:27, 10 November 2024
Contents
Problem
One side of an equilateral triangle of height lies on line . A circle of radius is tangent to line and is externally tangent to the triangle. The area of the region exterior to the triangle and the circle and bounded by the triangle, the circle, and line can be written as , where , , and are positive integers and is not divisible by the square of any prime. What is ?
Diagram
~MRENTHUSIASM
Solution 1
Call the bottom vertices and (the one closer to the circle is ) and the top vertex . The tangency point between the circle and the side of the triangle is , and the tangency point on line , and the center of the circle is
Draw radii to the tangency points, the arc is degrees because is , and since is supplementary, it's . The sum of the angles in a quadrilateral is , which means is
Triangle ODC is -- triangle so CD is .
Since we have congruent triangles ( and ), the combined area of both is .
The area of the arc is which is , so the answer is
is which is
~ASPALAPATI75
~andy_liu766 (latex)
edits by KR
Note
There were two possible configurations from this problem; the one described in the solution above and the configuration in which the circle is tangent to the bottom of line and the base of the equilateral triangle. However, since the area in this configuration is simply we can infer that the problem is talking about the configuration in Solution 1.
~dbnl
Solution 2 (Quick Guess)
Since this problem involves equilateral triangles, the only possible number under the square root is . Now subtracting all of the answer choices by , we get:
Due to the even parity of the problem, we can safely assume that the answer is either or , but as is a multiple of and , we get the answer of .
~megaboy6679
Solution 3
(pardon the diagrams :D)
say the area we want to find is x.
since the equilateral triangle has an internal angle of 60, the exterior angle formed by the triangle and the line is 120. simplifying the diagram you will get:
\ ##### \ ######## \######## \#####___________
make three of these that each circle is tangent to the other 2 circles
\ \ ##### \ ######## #####\######## #######\_####__________ #######/ ##### #####/######## / ######## / ######
Since they are 3 congruent triangles, you can make an equilateral triangle using their radius(12), with each vertex at the center of each circle. This will make an equilateral triangle of side length 24. if you look now, the area within the equilateral triangle consists of 3 of a circle, and 3 of x.
we first find the area of the triangle, which is , we then find the area of of a circle, which is , we subtract by , and divide by 3, yielding the area of x.
-
_________________________________________ =
3
is which is
~Yiguo Zhang
Video Solution by Pi Academy
https://youtu.be/ABkKz0gS1MU?si=ZQBgDMRaJmMPSSMM
Video Solution 1 by Power Solve
See also
2024 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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.