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# Geometry Set

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.

1. 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)

2. 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 $$AQ=D_2P$$.

3. 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?

Note by José Marín Guzmán
2 years, 1 month ago

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3)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 $$n=7$$. · 2 years, 1 month ago

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