Rectangle $ABCD$ has sides $AB = 3$, $BC = 2$. Point $ P$ lies on side $AB$ is such that the bisector of the angle $CDP$ passes through the midpoint $M$ of $BC$. Find $BP$.
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Tags: geometry, rectangle
Arrowhead575
16.09.2021 17:24
I used Trig. Let $AP=k$ Let $<CDM=\theta$, then $sin(\theta)=1/\sqrt{10}, cos(\theta)=3/\sqrt{10}$, however we also know $cos(2*\theta)=\frac{k}{\sqrt{4+k^2}}$, giving $ k=8/3$, then $BP=3-8/3=1/3.$
Albert123
16.09.2021 17:31
Using Trigonometric: Let $M$ is midpoint of $BC$ and $BP=a$ , $AP=3-a$ Let $\angle PDM = \angle MDC = \alpha$ In triangle $MCD$: $tan(\alpha)=1/3$ In triangle $APD$: $tan( 2\alpha)=2/(3-x)$ Then, $x=1/3$
OlympusHero
16.09.2021 20:34
YUH.
Denote $\angle MDC = x$. Then $\sin x = \frac{1}{\sqrt{10}}$ and $\cos x = \frac{3}{\sqrt{10}}$, so $2 \sin x \cos x = \sin(2x)=\cos(90-2x)=\frac{2}{DP}=\frac{3}{5}$. This means $DP=\frac{10}{3}$, so $AP=\frac{8}{3}$ and $BP=3-\frac{8}{3} = \boxed{\frac{1}{3}}$.
NealShrestha
16.09.2021 20:44
OlympusHero wrote: YUH.
Denote $\angle MDC = x$. Then $\sin x = \frac{1}{\sqrt{10}}$ and $\cos x = \frac{3}{\sqrt{10}}$, so $2 \sin x \cos x = \sin(2x)=\cos(90-2x)=\frac{2}{DP}=\frac{3}{5}$. This means $DP=\frac{10}{3}$, so $AP=\frac{8}{3}$ and $BP=3-\frac{8}{3} = \boxed{\frac{1}{3}}$.
Why post the exact same solution as #2? Like how does it add to the discussion?