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
2018 AMC 12A-Cyclic Quadrilaterals
contest-math
Related Solutions
Let P be the midpoint of FB. Extend MP to cut TF at Q.
After such construction, MBQF is a kite with $\angle DFB = \angle DFQ = \angle FBQ$ and $\angle BQM = \angle FQM$. The latter implies QM is the external angle bisector of $\angle BQF$ of $\triangle TQB$.
Draw QR // AB cutting TB at R. Since $\angle RQM = \angle BPM = 90^0$, QR is then the internal angle bisector of $\angle TQB$.
Therefore, $\dfrac {TM}{BM} = \dfrac {TR}{RB}$.
But, $\dfrac {TR}{RB}= \dfrac {TQ}{QF}$; because of similar triangles.
Further, $\dfrac {TQ}{QF} = \dfrac {TB}{BD}$; because of similar triangles. .
Using the given data, we can write, $$ \begin{matrix} TG=TA-AG=TA-1 & \mathrm{and} & TU=TP-PU=TP-1 \\ \end{matrix} .$$
Since $B$ and $M$ are the midpoints of $AT$ and $AP$ respectively, $BM$ is parallel to $TP$. Similarly, $CN$ is parallel to $TP$, because $N$ and $C$ are the midpoints of $UG$ and $TG$ respectively. Therefore, we also aware that, $$ \begin{matrix} BM=\frac{TP}{2} & \mathrm{and} & CN=\frac{TU}{2}=\frac{TP}{2}-\frac{1}{2} \\ \end{matrix} .$$ Now, we can write, $$BM-CN=\frac{TP}{2}-\left(\frac{TP}{2}-\frac{1}{2}\right)=\frac{1}{2}\tag{1}$$
We also know, that, $$ \begin{matrix} AB=\frac{AT}{2} & \mathrm{and} & GC=\frac{GT}{2}=\frac{AT}{2}-\frac{1}{2}=AB-\frac{1}{2} \\ \end{matrix} .$$ Since $BG=AB-AG=AB-1$, we have, $$BC=CG-BG=\frac{1}{2}.$$
Since $ND//BC$ and $NC//DB$, we have $$ \begin{matrix} BD=CN & \mathrm{and} & DN=BC=\frac{1}{2} \\ \end{matrix} .$$
To find $DM$, we write, $$MD=BM-BD=\left(CN+\frac{1}{2}\right)-CN=\frac{1}{2}=DN.\tag{2}$$
Since $ND//BC$, $\measuredangle ABD$ and $\measuredangle BDN$ are alternate angles. Therefore, they are equal, i.e. $$\measuredangle ABD=\measuredangle BDN \tag{3}.$$
Since $MD=DN$, $MDN$ is an isosceles triangle. Therefore, $$\measuredangle DMN=\frac{180^0-\measuredangle ABD}{2}=\frac{180^0-\measuredangle ATP}{2}=\frac{180^0-56^0 - 36^0}{2}=44^0.$$
Finally, we have, $$\measuredangle NMA=\measuredangle NMD + \measuredangle BMA = \measuredangle NMD + \measuredangle TPA=44^0+36^0=80^0.\tag{4}$$
Best Answer
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.