Difference between revisions of "2018 AMC 12A Problems/Problem 20"
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== 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>. Since <math>{AIME}</math> is cyclic, <math>\angle EMI</math> (which is opposite of another right angle) must be a right angle; therefore, <math>\vec{ME} \cdot \vec{MI} = <\frac{-3}{2}, y-\frac{3}{2}> \cdot <x-\frac{3}{2}, -\frac{3}{2}> = 0</math>. | + | 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>. Since <math>{AIME}</math> is cyclic, <math>\angle EMI</math> (which is opposite of another right angle) must be a right angle; therefore, <math>\vec{ME} \cdot \vec{MI} = <\frac{-3}{2}, y-\frac{3}{2}> \cdot <x-\frac{3}{2}, -\frac{3}{2}> = 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 <math>2=(\frac{1}{2})[(\frac{3}{2})(y-\frac{3}{2})+(x)(-y)+(x+\frac{3}{2})(\frac{3}{2})]</math>. 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)=</math> <math>(\frac{3 \pm \sqrt{7}}{2},\frac{3 \mp \sqrt{7}}{2})</math>. But <math>AI>AE</math>, meaning <math>x=AI=\frac{3 + \sqrt{7}}{2} \rightarrow CI = 3-\frac{3 + \sqrt{7}}{2}=\frac{3 - \sqrt{7}}{2}</math>, and the final answer is <math>3+7+2=\boxed{12}</math>. |
==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 00:34, 30 July 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 .
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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