KVPY 2016 geometry

I wiil use such solution which uses more formulas:

In the figure above, first complete the semicircle to form a circle.

Reflect $D$ in the line $AB$ to get $F$ on the circle. So, let $DC = CF = x$ .

Now by intersecting chord theorem ,

$x^2 = (\frac23)(\frac43)$ hence, $DC = \frac{2\sqrt2}{3}$ .

Now, Call by $\theta$ , the angle $AOD$ and by $\alpha$ , the angle $ODE$ .

In triangle $DCO$ ,

$\cos { \theta } =\frac { 1 }{ 3 }$ . And by letting $AD = y$ and applying cosine rule in same triangle, we get

$\huge\ \frac { 1 }{ 3 } = \frac { 1+1-{ \left( AD \right) }^{ 2 } }{ 2 } \\ \Rightarrow AD = \frac { 2 }{ \sqrt { 3 } }$

So that $\large\ DE = \frac { 1 }{ \sqrt { 3 } }$ .

Now, in triangle $DOE$ ,

$\cos { \alpha =\frac { 1 }{ \sqrt { 3 } } }$ $...(1)$

We know that $DH = 2R\cos { \alpha }$ , where $R$ is circumradius of $AOD$ .

From, $\large\ R=\frac { abc }{ 4\triangle }$ , we get by using $(1)$

$\huge\ R = \frac { abc }{ 4\triangle }=\frac { \frac { 2 }{ \sqrt { 3 } } }{ 4\times \frac { 1 }{ 2 } \times 1\times 1\times \frac { 2\sqrt { 2 } }{ 3 } } =\frac { 3 }{ 2\sqrt { 6 } }$ .

So, $\large\ DH = \frac { 1 }{ \sqrt { 2 } }$ .

First, drop a perpendicular from $E$ onto $AB$ , intersecting $AB$ at point $F$ . Now, notice that $E$ is the midpoint of $AD$ due to the property that the perpendicular to a chord always bisects it. Since $EF$ is parallel to $DC$ by construction, $F$ is also the midpoint of $AH$ . Additionally $EF = \frac{1}{2}DC$ .

Thus $AF = FC = CO = \frac{1}{3}$ . This means that $C$ is also the midpoint of $FO$ . By construction again, $EF$ is parallel to $HC$ , hence $HC = \frac{1}{2}EF$ .

Therefore, $DH = DC - HC = DC - \frac{1}{2}EF = DC - \frac{1}{4} DC= \frac{3}{4} DC$ . Now, connect $DO$ and observe $DO = 1$ . Thus by pythagoras' theorem, $DC = \sqrt{1-\frac{1}{3}^2}=\sqrt{\frac{8}{9}}$ . Finally, we have $DH = \frac{3}{4}\sqrt{\frac{8}{9}}=0.707$ .

We first get $AC=\frac{2}{3},CO=\frac{1}{3}$ and $EC=\frac{2\sqrt2}{3}$ using Pythagoras Theorem on $\triangle ECO.$
We observe that $AL=LE$ since $\triangle AOE$ is isosceles and $CH=\frac{2\sqrt2-3EH}{3}.$
We apply Menelaus Theorem for $\triangle ECA$ with $LO$ as the transversal.
So, $\dfrac{AO}{CO}\times\dfrac{CH}{EH}\times\dfrac{LE}{AL}=1$ $[$ Actually it should be $-1$ but it doesn't affect the problem as the minus sign is just to indicate that the point $O$ does not lie between $AC$ $].$
$\Rightarrow\dfrac{1}{\frac{1}{3}}\times \dfrac{CH}{EH}\times\dfrac{\cancel{LE}}{\cancel{AL}}=1.$
$\Rightarrow \dfrac{CH}{EH}=\dfrac{1}{3}.$
$\Rightarrow \dfrac{2\sqrt2-3EH}{\cancel{3}EH}=\dfrac{1}{\cancel{3}}.$
Therefore after solving we get $EH=\boxed{\dfrac{1}{\sqrt2}}.$