Difference between revisions of "2018 AMC 12A Problems/Problem 20"
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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>. | ||
− | + | ==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 <math>(3-a)^2+(3-b)^2=8\implies (3-a)^2+a^2=8\implies 2a^2-6a+9=8\implies 2a^2-6a+1=0\implies a=\frac{6\pm \sqrt{36-8}}{2}=\frac{3\pm\sqrt{7}}{2}</math>. 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{12\textbf{(D)}}</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 13:40, 6 April 2020
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 ), 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 (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 .
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
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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