The problems are in no particular order. Please reshare the note if you like what you see, feel free to share comments or ideas and enjoy.

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?

## Comments

Sort by:

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 \(n=7\). – Souryajit Roy · 2 years, 7 months ago

Log in to reply