Difference between revisions of "2018 AMC 12A Problems/Problem 20"
(→Solution 4 (Coordinate Geometry)) |
MRENTHUSIASM (talk | contribs) m (→Solution 6 (Bash): Fixed the extremely long line of equation.) |
||
(16 intermediate revisions by 8 users not shown) | |||
Line 10: | Line 10: | ||
\textbf{(E) }13 \qquad | \textbf{(E) }13 \qquad | ||
</math> | </math> | ||
+ | |||
+ | == Diagram == | ||
+ | <center> | ||
+ | <asy> | ||
+ | import olympiad; | ||
+ | |||
+ | size(200); | ||
+ | |||
+ | pair A, B, C, I, M, E; | ||
+ | |||
+ | A = (0, 0); | ||
+ | B = (3, 0); | ||
+ | C = (0, 3); | ||
+ | M = (1.5, 1.5); | ||
+ | I = (0, 1.5 + sqrt(2) / 2); | ||
+ | E = (1.5 - sqrt(2) / 2, 0); | ||
+ | |||
+ | draw(A -- B -- C -- cycle); | ||
+ | draw(I -- M -- E -- cycle); | ||
+ | draw(rightanglemark(I, A, E, 4)); | ||
+ | |||
+ | dot(A); | ||
+ | dot(B); | ||
+ | dot(C); | ||
+ | dot(I); | ||
+ | dot(M); | ||
+ | dot(E); | ||
+ | |||
+ | label("$A$", A, SW); | ||
+ | label("$B$", B, E); | ||
+ | label("$C$", C, N); | ||
+ | label("$I$", I, NE); | ||
+ | label("$M$", M, NE); | ||
+ | label("$E$", E + (0.1, 0.04), NE); | ||
+ | label("$3$", (A + C) / 2, W); | ||
+ | label("$3$", (A + B) / 2, S); | ||
+ | </asy> | ||
+ | </center> | ||
== Solution 1== | == Solution 1== | ||
− | Observe that <math>\triangle{EMI}</math> is isosceles right (<math>M</math> is the midpoint of diameter arc <math>EI</math>), so <math>MI=2,MC=\frac{3}{\sqrt{2}}</math>. With <math>\angle{MCI}=45^\circ</math>, we can use Law of Cosines to determine that <math>CI=\frac{3\pm\sqrt{7}}{2}</math>. The same calculations hold for <math>BE</math> also, and since <math>CI<BE</math>, we deduce that <math>CI</math> is the smaller root, giving the answer of <math>\boxed{12}</math>. | + | Observe that <math>\triangle{EMI}</math> is isosceles right (<math>M</math> is the midpoint of diameter arc <math>EI</math> since <math>m\angle MEI = m\angle MAI = 45^\circ</math>), so <math>MI=2,MC=\frac{3}{\sqrt{2}}</math>. With <math>\angle{MCI}=45^\circ</math>, we can use Law of Cosines to determine that <math>CI=\frac{3\pm\sqrt{7}}{2}</math>. The same calculations hold for <math>BE</math> also, and since <math>CI<BE</math>, we deduce that <math>CI</math> is the smaller root, giving the answer of <math>\boxed{\textbf{(D) }12}</math>. |
− | |||
− | == Solution 2 ( | + | == Solution 2 (Ptolemy) == |
We first claim that <math>\triangle{EMI}</math> is isosceles and right. | We first claim that <math>\triangle{EMI}</math> is isosceles and right. | ||
Line 26: | Line 63: | ||
Since <math>M</math> is the mid-point of <math>\overline{BC}</math>, it is clear that <math>AM=\frac{3\sqrt{2}}{2}</math>. | Since <math>M</math> is the mid-point of <math>\overline{BC}</math>, it is clear that <math>AM=\frac{3\sqrt{2}}{2}</math>. | ||
− | Now let <math>AE=a</math> and <math>AI=b</math>. By Ptolemy's Theorem, in cyclic quadrilateral <math>AIME</math>, we have <math>2a+2b=6</math>. By Pythagorean Theorem, we have <math>a^2+b^2=8</math>. One can solve the simultaneous system and find <math>b=\frac{3+\sqrt{7}}{2}</math>. Then by deducting the length of <math>\overline{AI}</math> from 3 we get <math>CI=\frac{3-\sqrt{7}}{2}</math>, giving the answer of <math>\boxed{12}</math>. (Surefire2019) | + | Now let <math>AE=a</math> and <math>AI=b</math>. By Ptolemy's Theorem, in cyclic quadrilateral <math>AIME</math>, we have <math>2a+2b=6</math>. By Pythagorean Theorem, we have <math>a^2+b^2=8</math>. One can solve the simultaneous system and find <math>b=\frac{3+\sqrt{7}}{2}</math>. Then by deducting the length of <math>\overline{AI}</math> from 3 we get <math>CI=\frac{3-\sqrt{7}}{2}</math>, giving the answer of <math>\boxed{\textbf{(D) }12}</math>. (Surefire2019) |
− | == Solution 3 ( | + | == Solution 3 (Elementary) == |
− | Like above, notice that <math>\triangle{EMI}</math> is isosceles and right, which means that <math>\dfrac{ME \cdot MI}{2} = 2</math>, so <math>MI^2=4</math> and <math>MI = 2</math>. Then construct <math>\overline{MF}\perp\overline{AB}</math> and <math>\overline{MG}\perp\overline{AC}</math> as well as <math>\overline{MI}</math>. It's clear that <math>MG^2+GI^2 = MI^2</math> by Pythagorean, so knowing that <math>MG = \dfrac{AB}{2} = \dfrac{3}{2}</math> allows one to solve to get <math>GI = \dfrac{\sqrt{7}}{2}</math>. By just looking at the diagram, <math>CI=AC-MF-GI=\dfrac{3-\sqrt{7}}{2}</math>. The answer is thus <math>3+7+2=12</math>. | + | Like above, notice that <math>\triangle{EMI}</math> is isosceles and right, which means that <math>\dfrac{ME \cdot MI}{2} = 2</math>, so <math>MI^2=4</math> and <math>MI = 2</math>. Then construct <math>\overline{MF}\perp\overline{AB}</math> and <math>\overline{MG}\perp\overline{AC}</math> as well as <math>\overline{MI}</math>. It's clear that <math>MG^2+GI^2 = MI^2</math> by Pythagorean, so knowing that <math>MG = \dfrac{AB}{2} = \dfrac{3}{2}</math> allows one to solve to get <math>GI = \dfrac{\sqrt{7}}{2}</math>. By just looking at the diagram, <math>CI=AC-MF-GI=\dfrac{3-\sqrt{7}}{2}</math>. The answer is thus <math>3+7+2=\boxed{\textbf{(D) }12}</math>. |
== Solution 4 (Coordinate Geometry) == | == Solution 4 (Coordinate Geometry) == | ||
− | Let <math>A</math> lie on <math>(0,0)</math>, <math>E</math> on <math>(0,y)</math>, <math>I</math> on <math>(x,0)</math>, and <math>M</math> on <math>(\frac{3}{2},\frac{3}{2})</math>. <math>{AIME}</math> is cyclic, | + | Let <math>A</math> lie on <math>(0,0)</math>, <math>E</math> on <math>(0,y)</math>, <math>I</math> on <math>(x,0)</math>, and <math>M</math> on <math>\left(\frac{3}{2},\frac{3}{2}\right)</math>. Since <math>{AIME}</math> is cyclic, <math>\angle EMI</math> (which is opposite of another right angle) must be a right angle; therefore, <math>\overrightarrow{ME} \cdot \overrightarrow{MI} = \left<\frac{-3}{2}, y-\frac{3}{2}\right> \cdot \left<x-\frac{3}{2}, -\frac{3}{2}\right> = 0</math>. Compute the dot product to arrive at the relation <math>y=3-x</math>. We can set up another equation involving the area of <math>\triangle EMI</math> using the [[Shoelace Theorem]]. This is <cmath>2=\frac{1}{2}\left[\frac{3}{2}\left(y-\frac{3}{2}\right)-xy+\frac{3}{2}\left(x+\frac{3}{2}\right)\right].</cmath> Multiplying, substituting <math>3-x</math> for <math>y</math>, and simplifying, we get <math>x^2 -3x + \frac{1}{2}=0</math>. Thus, <math>(x,y)=\left(\frac{3 \pm \sqrt{7}}{2},\frac{3 \mp \sqrt{7}}{2}\right)</math>. But <math>AI>AE</math>, meaning <math>x=AI=\frac{3 + \sqrt{7}}{2}</math> and <math>CI = 3-\frac{3 + \sqrt{7}}{2}=\frac{3 - \sqrt{7}}{2}</math>, and the final answer is <math>3+7+2=\boxed{\textbf{(D) }12}</math>. |
+ | |||
+ | == Solution 5 (Quick) == | ||
+ | From <math>AIME</math> cyclic we get <math>\angle{MEI} = \angle{MAI} = 45^\circ</math> and <math>\angle{MIE} = \angle{MAE} = 45^\circ</math>, so <math>\triangle{EMI}</math> is an isosceles right triangle. | ||
+ | |||
+ | From <math>[EMI]=2</math> we get <math>EM=MI=2</math>. | ||
+ | |||
+ | Notice <math>\triangle{AEM} \cong \triangle{CIM}</math>, because <math>\angle{AEM}=180-\angle{AIM}=\angle{CIM}</math>, <math>EM=IM</math>, and <math>\angle{EAM}=\angle{ICM}=45^\circ</math>. | ||
+ | |||
+ | Let <math>CI=AE=x</math>, so <math>AI=3-x</math>. | ||
+ | |||
+ | By Pythagoras on <math>\triangle{EAI}</math> we have <math>x^2+(3-x)^2=EI^2=8</math>, and solve this to get <math>x=CI=\dfrac{3-\sqrt{7}}{2}</math> for a final answer of <math>3+7+2=\boxed{\textbf{(D) }12}</math>. | ||
+ | |||
+ | ==Solution 6 (Bash)== | ||
+ | Let <math>CI=a</math>, <math>BE=b</math>. Because opposite angles in a cyclic quadrilateral are supplementary, we have <math>\angle EMI=90^{\circ}</math>. By the law of cosines, we have <math>MI^2=a^2+\frac{9}{4}-\frac{3}{2}a</math>, and <math>ME^2=b^2+\frac{9}{4}-\frac{3}{2}b</math>. Notice that <math>EI=2MO</math>, where <math>O</math> is the origin of the circle mentioned in the problem. Thus <math>\frac{2MO*MO}{2}=2\implies MO=\sqrt{2}, EI=2\sqrt{2}</math>. By the Pythagorean Theorem, we have <math>ME^2+MI^2=EI^2\implies a^2+\frac{9}{4}-\frac{3}{2}a+b^2+\frac{9}{4}-\frac{3}{2}b=(2\sqrt{2})^2=8</math>. By the Pythagorean Theorem, we have <math>AE^2+AI^2=EI^2\implies (3-a)^2+(3-b)^2=(2\sqrt{2})^2=8\implies 18-6a-6b+a^2+b^2=8</math>. Thus we have <math>18-6a-6b+a^2+b^2=a^2+\frac{9}{4}-\frac{3}{2}a+b^2+\frac{9}{4}-\frac{3}{2}b\implies 18-6a-6b=\frac{9}{2}-\frac{3}{2}a-\frac{3}{2}b\implies \frac{27}{2}</math> <math>=\frac{9}{2}a+\frac{9}{2}b\implies a+b=3\implies 3-b=a</math>. We know that | ||
+ | <cmath>\begin{align*} | ||
+ | (3-a)^2+(3-b)^2&=8 \\ | ||
+ | (3-a)^2+a^2&=8 \\ | ||
+ | 2a^2-6a+9&=8 \\ | ||
+ | 2a^2-6a+1&=0 \\ | ||
+ | a&=\frac{6\pm \sqrt{36-8}}{2}=\frac{3\pm\sqrt{7}}{2}. | ||
+ | \end{align*}</cmath> | ||
+ | We take the smaller solution because we have <math>AI>AE\implies 3-AI<3-AE\implies CI<CE</math>, and we want <math>CI</math>, not <math>CE</math>, thus <math>CI=\frac{3-\sqrt{7}}{2}</math>. Thus our final answer is <math>3+7+2=\boxed{\textbf{(D) }12}</math>. | ||
+ | |||
+ | -vsamc | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/NsQbhYfGh1Q?t=4465 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2018|ab=A|num-b=19|num-a=21}} | {{AMC12 box|year=2018|ab=A|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 18:07, 19 August 2021
Contents
Problem
Triangle is an isosceles right triangle with . Let be the midpoint of hypotenuse . Points and lie on sides and , respectively, so that and is a cyclic quadrilateral. Given that triangle has area , the length can be written as , where , , and are positive integers and is not divisible by the square of any prime. What is the value of ?
Diagram
Solution 1
Observe that is isosceles right ( is the midpoint of diameter arc since ), so . With , we can use Law of Cosines to determine that . The same calculations hold for also, and since , we deduce that is the smaller root, giving the answer of .
Solution 2 (Ptolemy)
We first claim that is isosceles and right.
Proof: Construct and . Since bisects , one can deduce that . Then by AAS it is clear that and therefore is isosceles. Since quadrilateral is cyclic, one can deduce that . Q.E.D.
Since the area of is 2, we can find that ,
Since is the mid-point of , it is clear that .
Now let and . By Ptolemy's Theorem, in cyclic quadrilateral , we have . By Pythagorean Theorem, we have . One can solve the simultaneous system and find . Then by deducting the length of from 3 we get , giving the answer of . (Surefire2019)
Solution 3 (Elementary)
Like above, notice that is isosceles and right, which means that , so and . Then construct and as well as . It's clear that by Pythagorean, so knowing that allows one to solve to get . By just looking at the diagram, . The answer is thus .
Solution 4 (Coordinate Geometry)
Let lie on , on , on , and on . Since is cyclic, (which is opposite of another right angle) must be a right angle; therefore, . Compute the dot product to arrive at the relation . We can set up another equation involving the area of using the Shoelace Theorem. This is Multiplying, substituting for , and simplifying, we get . Thus, . But , meaning and , and the final answer is .
Solution 5 (Quick)
From cyclic we get and , so is an isosceles right triangle.
From we get .
Notice , because , , and .
Let , so .
By Pythagoras on we have , and solve this to get for a final answer of .
Solution 6 (Bash)
Let , . Because opposite angles in a cyclic quadrilateral are supplementary, we have . By the law of cosines, we have , and . Notice that , where is the origin of the circle mentioned in the problem. Thus . By the Pythagorean Theorem, we have . By the Pythagorean Theorem, we have . Thus we have . We know that We take the smaller solution because we have , and we want , not , thus . Thus our final answer is .
-vsamc
Video Solution
https://youtu.be/NsQbhYfGh1Q?t=4465
~ pi_is_3.14
See Also
2018 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.