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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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