On a cyclic quadrilateral $ABCD$, there is a point $P$ on side $AD$ such that the triangle $CDP$ and the quadrilateral $ABCP$ have equal perimeters and equal areas. Prove that two sides of $ABCD$ have equal lengths.
Problem
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Tags: geometry, cyclic quadrilateral, perimeter, PMO, 2010
Irrational_phi
16.11.2015 13:27
Let $AB=a, BC=b, CD=c, DP=d, PA=e$. Denote by $[a_1a_2\cdots a_n]$ the area of the $n$-gon $a_1a_2\cdots a_n$. Since $\triangle ACP$ and $ \triangle CPD$ share the same height from $C$, we have that $\frac{[ACP]}{[PCD]}=\frac{e}{d}$. Let $\angle ABC= \theta$. Since quadrilateral $ABCD$ is cyclic, $\angle CDA=\pi - \theta$. So $\sin(\angle ABC)=\sin(\angle CDA)$. So $\frac{[ABC]}{[CPD]}=\frac{\frac{1}{2}ab\sin \theta}{\frac{1}{2}cd\sin\theta}=\frac{ab}{cd}$. Since $[ABCP]=[PCD]$, $\frac{ab}{cd}+\frac{e}{d}=1 \implies c(d-e)-ab=0$. Since quadrilateral $ABCP$ and triangle $PCD$ have the same perimeter, $a+b+e=c+d \implies d-e=a+b-c$. Since $0=c(d-e)-ab=c(a+b-c)-ab=-(c-a)(c-b)$, we must have $c=b$ or $c=a$, and the conclusion follows.
drmzjoseph
17.11.2015 19:03
Posted Before geometry
Takeya.O
02.11.2016 05:14
Let $PD=a,PA=b,AB=c,BC=d,CD=e,\angle ADC=\theta$
Then
$\triangle CDP=\frac{1}{2}ae\sin \theta$
$\triangle ACP=\triangle CDP\cdot \frac{b}{a}=\frac{1}{2}be\sin \theta$
$\triangle ABC=\frac{1}{2}cd\sin (\pi-\theta)=\frac{1}{2}cd\sin \theta$
From condition,
$a+e=b+c+d$
$\frac{1}{2}ae\sin \theta=\frac{1}{2}be\sin \theta+\frac{1}{2}cd\sin \theta$
$\Leftrightarrow ae=be+cd\Leftrightarrow a=b+\frac{cd}{e}$
Substituting this into $a+e=b+c+d,$then
$\frac{cd}{e}+e=c+d\Leftrightarrow (c-e)(d-e)=0\implies c=e\text{ or }d=e$
Therefore the proof is completed.$\blacksquare$
rafaello
03.12.2020 22:41
By equal perimeters, $$PD+CD=PA+AB+BC\quad (1)$$Calculating area of $ABCD$ and $PDC$, knowing that $S_{PDC}=\frac{1}{2}\cdot S_{ABCD}$, we get that $$S_{ABCD}=AB\cdot BC\cdot \frac{1}{2}\cdot\sin{\angle ABC}+DA\cdot DC\cdot \frac{1}{2}\cdot\sin{\angle PDC}=\frac{1}{2}\cdot\sin{\angle ABC} (AB\cdot BC+DA\cdot DC)$$$$CD\cdot DP\cdot \frac{1}{2}\cdot\sin{\angle PDC}=S_{PDC}=\frac{1}{2}\cdot S_{ABCD}$$By the last two we obtain that $$2\cdot CD\cdot DP =AB\cdot BC+DA\cdot DC\implies AB\cdot BC+CD\cdot AP=DP\cdot DC$$Multiplying $(1)$ by $CD$, we get $$AB\cdot BC+CD\cdot AP=PA\cdot DC+AB\cdot DC+BC\cdot DC-CD\cdot DC\implies AB\cdot BC+CD^2=AB\cdot DC+BC\cdot DC$$The last is equivalent to $$BC\cdot (AB-CD)= DC\cdot (AB-CD),$$thus $AB=CD$ or $BC=CD$. We are done!