2018 AMC 12A-Cyclic Quadrilaterals
Triangle $ABC$ is an isosceles right triangle with $AB=AC=3$. Let $M$ be the midpoint of hypotenuse $overline{BC}$. Points $I$ and $E$ lie on sides $overline{AC}$ and $overline{AB}$, respectively, so that $AI>AE$ and $AIME$ is a cyclic quadrilateral. Given that triangle $EMI$ has area $2$, the length $CI$ can be written as $frac{a-sqrt{b}}{c}$, where $a$, $b$, and $c$ are positive integers and $b$ is not divisible by the square of any prime. What is the value of $a+b+c$?
How do we prove that triangle EMI is isosceles. Currently, I know that angle EMI is a right angle
contest-math
asked Dec 28 '18 at 1:05
user501887
183
add a comment |
Triangle $ABC$ is an isosceles right triangle with $AB=AC=3$. Let $M$ be the midpoint of hypotenuse $overline{BC}$. Points $I$ and $E$ lie on sides $overline{AC}$ and $overline{AB}$, respectively, so that $AI>AE$ and $AIME$ is a cyclic quadrilateral. Given that triangle $EMI$ has area $2$, the length $CI$ can be written as $frac{a-sqrt{b}}{c}$, where $a$, $b$, and $c$ are positive integers and $b$ is not divisible by the square of any prime. What is the value of $a+b+c$?
How do we prove that triangle EMI is isosceles. Currently, I know that angle EMI is a right angle
contest-math
asked Dec 28 '18 at 1:05
user501887
183
add a comment |
Triangle $ABC$ is an isosceles right triangle with $AB=AC=3$. Let $M$ be the midpoint of hypotenuse $overline{BC}$. Points $I$ and $E$ lie on sides $overline{AC}$ and $overline{AB}$, respectively, so that $AI>AE$ and $AIME$ is a cyclic quadrilateral. Given that triangle $EMI$ has area $2$, the length $CI$ can be written as $frac{a-sqrt{b}}{c}$, where $a$, $b$, and $c$ are positive integers and $b$ is not divisible by the square of any prime. What is the value of $a+b+c$?
How do we prove that triangle EMI is isosceles. Currently, I know that angle EMI is a right angle
contest-math
asked Dec 28 '18 at 1:05
user501887
183
Triangle $ABC$ is an isosceles right triangle with $AB=AC=3$. Let $M$ be the midpoint of hypotenuse $overline{BC}$. Points $I$ and $E$ lie on sides $overline{AC}$ and $overline{AB}$, respectively, so that $AI>AE$ and $AIME$ is a cyclic quadrilateral. Given that triangle $EMI$ has area $2$, the length $CI$ can be written as $frac{a-sqrt{b}}{c}$, where $a$, $b$, and $c$ are positive integers and $b$ is not divisible by the square of any prime. What is the value of $a+b+c$?
How do we prove that triangle EMI is isosceles. Currently, I know that angle EMI is a right angle
contest-math
contest-math
asked Dec 28 '18 at 1:05
user501887
183
asked Dec 28 '18 at 1:05
user501887
183
asked Dec 28 '18 at 1:05
user501887
183
asked Dec 28 '18 at 1:05
user501887
183
asked Dec 28 '18 at 1:05
user501887
183
183
add a comment |
add a comment |
1 Answer
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The fact that $triangle EMI$ is isosceles, namely $IM = EM$, is a direct consequence of the fact that $angle EMI = 90^circ$ and $M$ is the midpoint of $overline{BC}$. To see this, all one needs to do is consider a rotation of the entire figure about point $M$ by $90$ degrees, either clockwise or counterclockwise. Doing so will identify $triangle AMB$ with $triangle AMC$; moreover, it also identifies $E$ with $I$, since $angle EMI = 90^circ$ as previously stated. Hence $IM = EM$ as claimed.
answered Dec 28 '18 at 1:24
heropup
62.6k66099
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The fact that $triangle EMI$ is isosceles, namely $IM = EM$, is a direct consequence of the fact that $angle EMI = 90^circ$ and $M$ is the midpoint of $overline{BC}$. To see this, all one needs to do is consider a rotation of the entire figure about point $M$ by $90$ degrees, either clockwise or counterclockwise. Doing so will identify $triangle AMB$ with $triangle AMC$; moreover, it also identifies $E$ with $I$, since $angle EMI = 90^circ$ as previously stated. Hence $IM = EM$ as claimed.
answered Dec 28 '18 at 1:24
heropup
62.6k66099
add a comment |
The fact that $triangle EMI$ is isosceles, namely $IM = EM$, is a direct consequence of the fact that $angle EMI = 90^circ$ and $M$ is the midpoint of $overline{BC}$. To see this, all one needs to do is consider a rotation of the entire figure about point $M$ by $90$ degrees, either clockwise or counterclockwise. Doing so will identify $triangle AMB$ with $triangle AMC$; moreover, it also identifies $E$ with $I$, since $angle EMI = 90^circ$ as previously stated. Hence $IM = EM$ as claimed.
answered Dec 28 '18 at 1:24
heropup
62.6k66099
add a comment |
The fact that $triangle EMI$ is isosceles, namely $IM = EM$, is a direct consequence of the fact that $angle EMI = 90^circ$ and $M$ is the midpoint of $overline{BC}$. To see this, all one needs to do is consider a rotation of the entire figure about point $M$ by $90$ degrees, either clockwise or counterclockwise. Doing so will identify $triangle AMB$ with $triangle AMC$; moreover, it also identifies $E$ with $I$, since $angle EMI = 90^circ$ as previously stated. Hence $IM = EM$ as claimed.
answered Dec 28 '18 at 1:24
heropup
62.6k66099
The fact that $triangle EMI$ is isosceles, namely $IM = EM$, is a direct consequence of the fact that $angle EMI = 90^circ$ and $M$ is the midpoint of $overline{BC}$. To see this, all one needs to do is consider a rotation of the entire figure about point $M$ by $90$ degrees, either clockwise or counterclockwise. Doing so will identify $triangle AMB$ with $triangle AMC$; moreover, it also identifies $E$ with $I$, since $angle EMI = 90^circ$ as previously stated. Hence $IM = EM$ as claimed.
answered Dec 28 '18 at 1:24
heropup
62.6k66099
answered Dec 28 '18 at 1:24
heropup
62.6k66099
answered Dec 28 '18 at 1:24
heropup
62.6k66099
answered Dec 28 '18 at 1:24
heropup
62.6k66099
62.6k66099
add a comment |
add a comment |
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