A problem concerning a parallelogram and a circle

Let $ x =AL = AK$ then $DL = 20-x$ and $BK = 16-x$. Also $BM = 20/9$

By the PoP with respect to $B$ we have $$BM\cdot BC = BK^2\implies BK = 20/3\implies x= 28/3$$

By the PoP with respect to $D$ we have $$DN\cdot DA = DL^2\implies DL = ...$$


We have to be careful because the given picture is misleading, actually $K$ and $L$ lie outside $ABCD$.

enter image description here

We have $BM\cdot BC = \frac{1}{9}BC^2 = \frac{400}{9}=BK^2$, hence $BK=\frac{20}{3}$ and $AL=AK=16+\frac{20}{3}=\frac{68}{3}$, such that $DL=\frac{8}{3}$. This gives $DN\cdot DC=\frac{64}{9}$, hence $DN=\frac{4}{9}$ and $\frac{DN}{NC}=\frac{4/9}{16-4/9}=\color{red}{\frac{1}{35}}$.

There is a second solution with $\widehat{DAB}\approx 96.38^\circ$ and $\frac{DN}{NC}=\color{red}{\frac{4}{5}}$; in this case $K$ and $L$ properly lie on $AB$ and $AD$. This is probably the intended solution if we label the vertices of $ABCD$ counter-clockwise, as usually done.


Different from power of a point solution:
Vectors.
Let $b:=\overrightarrow{AB}$, $\;d:=\overrightarrow{AD}$, $c:=\cos\angle BAD$.
It's given that $|\overrightarrow{AB}|=16$, $\;|\overrightarrow{AD}|=20$, $\;\overrightarrow{AM}=b+\frac19 d$ and $\rho(O,AB)$ $=\rho(O,AD)$ $=OC$ $=OM$.
$\rho(O,AB)=\rho(O,AD)$ implies $AO$ is the bisector of $\angle BAD$ thus $\overrightarrow{AO}||\left( \frac{\overrightarrow{AB}}{|\overrightarrow{AB}|}+ \frac{\overrightarrow{AD}}{|\overrightarrow{AD}|} \right)$.

Let $\overrightarrow{AO}=t(5b+4d)$ for some $t$. We find $\rho(O,AD)=\sqrt{(\overrightarrow{AO})^2-\left(b \dfrac{\overrightarrow{AO}\cdot b}{|b|^2}\right)^2}$

Then we have $$ \begin{cases} (t(5b+4d)-(b+d))^2=(t(5b+4d)-(b+\frac19 d))^2\\ (t(5b+4d)-(b+d))^2=(t(5b+4d))^2-\frac{1}{b^2}\left(t(5b+4d)\cdot b\right)^2 \end{cases} $$ Having $b^2=16^2$, $\;d^2=20^2$, $\;bd=16\cdot 20\cdot c$ we feed this thing to wolframalpha(1, 2, 3), obtaining $$ \begin{cases} (180 c + 180) t = 36 c + 25\\ (10 c^2 + 10) t^2 + c (20 t^2 - 9 t + 1) - 9 t = -\frac{41}{40} \end{cases} $$ $$ \left[ \begin{array}{l} \begin{cases} c = -\frac{1}{9}\\ t = \frac{21}{160} \end{cases}\\ \begin{cases} c = \frac{13}{18}\\ t = \frac{51}{310} \end{cases} \end{array} \right. $$ Now let $\overrightarrow{AN}=ub+d$, with $(\overrightarrow{OC})^2=(\overrightarrow{ON})^2$ $$(t(5b+4d)-(b+d))^2=(t(5b+4d)-(ub+d))^2$$ which gives (4, 5, 6) $u=\frac{4}{9}$ in the first case ($\frac{DN}{NC}=\frac{u}{1-u}=\frac{4}{5}$) and $u=\frac{1}{36}$ ($\frac{DN}{NC}=\frac{u}{1-u}=\frac{1}{35}$) in the second case almost for free, alone to say the computations are almost unbearable by hand)

Tags:

Geometry

Euclidean Geometry

Triangles

Circles

Power Of The Point

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