Geometry Set

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  1. Let PP and QQ be points on the side ABAB of the triangle ABCABC (with PP between AA and QQ) such that ACP=PCQ=QCB\angle{ACP}=\angle{PCQ}=\angle{QCB}, and let ADAD be the angle bisector of BAC\angle{BAC} with DD on BCBC. Line ADAD meets lines CPCP and CQCQ at MM and NN respectively. Given that PN=CDPN=CD and 3BAC=2BCA3\angle{BAC}=2\angle{BCA}, prove that triangles CQDCQD and QNBQNB have the same area. (Belarus 1999)

  2. Let ABCABC be a triangle and let ω\omega be its incircle. Let D1D_1 and E1E_1 be the tangency points of ω\omega with sides BCBC and ACAC, respectively. Let D2D_2 and E2E_2 be points on sides BCBC and ACAC, respectively, such that CD2=BD1CD_2=BD_1 and CE2=AE1CE_2=AE_1, and let PP be the intersection point of segments AD2AD_2 and BE2BE_2. Circle ω\omega intersects segment AD2AD_2 in two points, being the closest one to AA denoted by QQ. Prove that AQ=D2PAQ=D_2P.

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

Note by José Marín Guzmán
4 years, 6 months 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 nn- sided.

Suppose the vertex after DD be EE.Hence the 55 consecutive vertices are A,B,C,D,EA,B,C,D,E ( arranged anticlockwise from A to E,say)

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


One can easily show that quadrilateral ACDEACDE is cyclic.

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

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



Since,these two diagonals are equal,number of vertices between AA and EE(moving clockwise) equals the number of vertices between AA and DD(moving anti-clockwise).

Hence,n5=2n-5=2 or n=7n=7.

Souryajit Roy - 4 years, 6 months ago

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