Difference between revisions of "2018 AMC 12A Problems/Problem 20"
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== Solution 5 (Quick) == | == 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>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. | ||
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From <math>[EMI]=2</math> we get <math>EM=MI=2</math>. | From <math>[EMI]=2</math> we get <math>EM=MI=2</math>. | ||
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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>. | 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>. | ||
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Let <math>CI=AE=x</math>, so <math>AI=3-x</math>. | Let <math>CI=AE=x</math>, so <math>AI=3-x</math>. | ||
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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{12}</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{12}</math>. | ||
Revision as of 01:33, 24 December 2018
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 ?
Solution 1
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)
Solution 3 (More 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 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 .
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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