Difference between revisions of "2018 AMC 12A Problems/Problem 20"
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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>. (trumpeter) | 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>. (trumpeter) | ||
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+ | == Solution 2 (Using Ptolemy) == | ||
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+ | We first claim that <math>\triangle{EMI}</math> is isosceles and right. | ||
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+ | Proof: Construct <math>\overline{MF}\perp\overline{AB}</math> and <math>\overline{MG}\perp\overline{AC}</math>. Since <math>\overline{AM}</math> bisects <math>\angle{BAC}</math>, one can deduce that <math>MF=MG</math>. Then by AAS it is clear that <math>MI=ME</math> and therefore <math>\triangle{EMI}</math> is isosceles. Since quadrilateral <math>AIME</math> is cyclic, one can deduce that <math>\angle{EMI}=90^\circ</math>. Q.E.D. | ||
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+ | Since the area of <math>\triangle{EMI}</math> is 2, we can find that <math>MI=ME=2</math>, <math>EI=2\sqrt{2}</math> | ||
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+ | 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>. | ||
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+ | 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) | ||
==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}} |
Revision as of 02:45, 9 February 2018
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 ?
Solution
Observe that is isosceles right ( is the midpoint of diameter arc ), 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 . (trumpeter)
Solution 2 (Using 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)
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 |
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