Difference between revisions of "2020 AMC 12A Problems/Problem 24"
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<math>\textbf{(A) } 1+\sqrt{2} \qquad \textbf{(B) } \sqrt{7} \qquad \textbf{(C) } \frac{8}{3} \qquad \textbf{(D) } \sqrt{5+\sqrt{5}} \qquad \textbf{(E) } 2\sqrt{2}</math> | <math>\textbf{(A) } 1+\sqrt{2} \qquad \textbf{(B) } \sqrt{7} \qquad \textbf{(C) } \frac{8}{3} \qquad \textbf{(D) } \sqrt{5+\sqrt{5}} \qquad \textbf{(E) } 2\sqrt{2}</math> | ||
− | ==Solution 1== | + | ==Solution== |
+ | |||
+ | <asy> | ||
+ | draw((0,0)--(4,5.65)--(8,0)--cycle); | ||
+ | label("A", (4,5.65), N, p = fontsize(10pt)); | ||
+ | label("C", (8,0), SE, p = fontsize(10pt)); | ||
+ | label("B", (0,0), SW, p = fontsize(10pt)); | ||
+ | label("P", (3.5,3.5), NW, p = fontsize(10pt)); | ||
+ | draw((0,0)--(3.5,3.5)); | ||
+ | label("$\sqrt{3}$",(0,0)--(3.5,3.5), SE); | ||
+ | draw((8,0)--(3.5,3.5)); | ||
+ | label("2",(8,0)--(3.5,3.5), SW); | ||
+ | draw((4,5.65)--(3.5,3.5)); | ||
+ | label("1",(4,5.65)--(3.5,3.5), E); | ||
+ | draw((8,0)--(4,5.65)--(11,5.65)--cycle); | ||
+ | label("$P'$", (6,4.75), NE, p = fontsize(10pt)); | ||
+ | draw((4,5.65)--(6,4.75)); | ||
+ | label("1",(4,5.65)--(6,4.75), S); | ||
+ | draw((8,0)--(6,4.75)); | ||
+ | label("$\sqrt{3}$",(8,0)--(6,4.75), E); | ||
+ | draw((3.5,3.5)--(6,4.75)); | ||
+ | label("1", (3.5,3.5)--(6,4.75), SE); | ||
+ | |||
+ | </asy> | ||
+ | |||
+ | We begin by rotating <math>\triangle{ABC}</math> by <math>60^{\circ}</math> about <math>A</math>, such that in <math>\triangle{A'B'C'}</math>, <math>B' = C</math>. We see that | ||
+ | <math>\triangle{APP'}</math> is equilateral with side length <math>1</math>, meaning that <math>\angle APP' = 60^{\circ}</math>. We also see that <math>\triangle{CPP'}</math> is a <math>30-60-90</math> right triangle, meaning that <math>\angle CPP'= 60^{\circ}</math>. Thus, by adding the two together, we see that <math>\angle APC = 120^{\circ}</math>. We can now use the law of cosines as following: | ||
+ | |||
+ | <cmath>s^2 = (AP)^2 + (CP)^2 - 2(AP)(CP)\cos{\angle{APC}}</cmath> | ||
+ | <cmath>s^2 = 1 + 4 - 2(1)(2)\cos{120^{\circ}}</cmath> | ||
+ | <cmath>s^2 = 5 - 4(-\frac{1}{2})</cmath> | ||
+ | <cmath>s = \sqrt{5 + 2}</cmath> | ||
+ | |||
+ | giving us that <math>s = \boxed{\textbf{(B) } \sqrt{7}}</math>. ~ciceronii |
Revision as of 19:31, 1 February 2020
Problem 24
Suppose that is an equilateral triangle of side length
, with the property that there is a unique point
inside the triangle such that
,
, and
. What is
?
Solution
We begin by rotating by
about
, such that in
,
. We see that
is equilateral with side length
, meaning that
. We also see that
is a
right triangle, meaning that
. Thus, by adding the two together, we see that
. We can now use the law of cosines as following:
giving us that . ~ciceronii