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Let $P$ and $Q$ be points on the side $AB$ of the triangle $ABC$ (with $P$ between $A$ and $Q$) such that $\angle{ACP}=\angle{PCQ}=\angle{QCB}$, and let $AD$ be the angle bisector of $\angle{BAC}$ with $D$ on $BC$. Line $AD$ meets lines $CP$ and $CQ$ at $M$ and $N$ respectively. Given that $PN=CD$ and $3\angle{BAC}=2\angle{BCA}$, prove that triangles $CQD$ and $QNB$ have the same area. (Belarus 1999)

Let $ABC$ be a triangle and let $\omega$ be its incircle. Let $D_1$ and $E_1$ be the tangency points of $\omega$ with sides $BC$ and $AC$, respectively. Let $D_2$ and $E_2$ be points on sides $BC$ and $AC$, respectively, such that $CD_2=BD_1$ and $CE_2=AE_1$, and let $P$ be the intersection point of segments $AD_2$ and $BE_2$. Circle $\omega$ intersects segment $AD_2$ in two points, being the closest one to $A$ denoted by $Q$. Prove that $AQ=D_2P$.

Points $A, B, C, D$ are four consecutive vertices of a regular polygon such that $\dfrac{1}{AB}=\dfrac{1}{AC}+\dfrac{1}{AD}$. How many sides does the polygon have?

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## Comments

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TopNewest3)The question number 3 is easy to solve with complex numbers.But also found a geometric solution.

Let the polygon be $n$- sided.

Suppose the vertex after $D$ be $E$.Hence the $5$ consecutive vertices are $A,B,C,D,E$ ( arranged anticlockwise from A to E,say)

Firstly,$\frac{1}{AB}=\frac{1}{AC}+\frac{1}{AD}$ gives

$AC.AD=AB.AD+AB.AC$...$(1)$

One can easily show that quadrilateral $ACDE$ is cyclic.

So,by Ptolemy's theorem, $AE.CD+AC.ED=AD.EC$...$(2)$

Also,since the polygon is regular,we have $AB=CD=ED$ and $EC=AC$.

So,from $(1)$, $AB.AD+AB.AC=AC.AD=CE.AD=AE.CD+AC.ED=AE.AB+AC.AB$

Hence,$AD=AE$.

Since,these two diagonals are equal,number of vertices between $A$ and $E$(moving clockwise) equals the number of vertices between $A$ and $D$(moving anti-clockwise).

Hence,$n-5=2$ or $n=7$.

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