Geometry (1st math Thailand POSN 2014)

1.) Explain how to construct a square with the same area of any given quadrilaterals.

2.) Let PP be a point in ABC\triangle ABC. Extend AP,BP,CP\overline{AP}, \overline{BP}, \overline{CP} intersect BC,CA,AB\overline{BC}, \overline{CA}, \overline{AB} at point D,E,FD,E,F respectively. If ACDF\square ACDF and BCEF\square BCEF are cyclic quadrilaterals, prove that PDBF\square PDBF is also cyclic quadrilateral.

3.) Use Ceva's theorem to prove that the 3 angle bisectors of ABC\triangle ABC intersect at 1 point.

4.) Let ABC\triangle ABC with point II as incenter. Extend AI\overline{AI} intersect circumcircle of ABC\triangle ABC at point PP.

  • 4.1) If point IaI_{a} is the center of excircle of ABC\triangle ABC which is opposite to angle AA, prove that PP is the center of the circumcenter of ICIaB\square ICI_{a}B

  • 4.2) Prove that 2sin(BI^C)=BCPI\displaystyle 2\sin{(B\hat{I}C)} = \frac{BC}{PI}

5.) Let PQRS\square PQRS has an incircle, which is tangent to SP,PQ,QR,RS\overline{SP}, \overline{PQ}, \overline{QR}, \overline{RS} at point A,B,C,DA,B,C,D respectively. Extend AB\overline{AB} intersect SQ\overline{SQ} at point XX. Prove that point C,D,XC,D,X are collinear.

This is the part of Thailand 1st round math POSN problems.

Note by Samuraiwarm Tsunayoshi
5 years ago

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1.) I tried to make the rectangle, let's say length a,ba,b. I tried to do by making a line length a+ba+b and draw a semicircle diameter of a+ba+b. Then I draw a perpendicular line at where the sides join and we get length ab\sqrt{ab}. And we're done, maybe.

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2) FDP=ADF=ACF\angle FDP=\angle ADF=\angle ACF since ACDFACDF is cyclic. ACF=ECP=FBP\angle ACF=\angle ECP=\angle FBP since BCEFBCEF is cyclic. Hence FBP=FDP    PDBF\angle FBP=\angle FDP \implies PDBF is cyclic.

Joel Tan - 5 years ago

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Problem 1: Easy one. See here.

Problem 2: First notice that PEF=BEF=BCF=DCF=DAF=PAF\angle PEF=\angle BEF=\angle BCF=\angle DCF=\angle DAF=\angle PAF so AFPEAFPE is cyclic. Hence DBF=CBF=AEF=APF=πFPD\angle DBF=\angle CBF=\angle AEF=\angle APF=\pi-\angle FPD implies PDBFPDBF is cyclic. Furthermore notice that PDB=ADB=πADC=πAFC=BFP\angle PDB=\angle ADB=\pi-\angle ADC=\pi-\angle AFC=\angle BFP so ADB=CFB=90\angle ADB=\angle CFB=90^\circ. So PP must be the orthocenter of ABC\triangle ABC.

Problem 3: Consider ABC\triangle ABC where ADAD, BEBE, CFCF are angle bisectors, with D,E,FD,E,F intersections with opposite sides. By Angle Bisector Theorem we have BD/DC=AB/ACBD/DC=AB/AC, CE/EA=BC/ABCE/EA=BC/AB, AF/FB=AC/BCAF/FB=AC/BC. Multiplying these cancels Ceva. \square

Problem 4: Check the problem, the wordings are wrong.

Problem 5: Menelaus on QRS\triangle QRS.

Jubayer Nirjhor - 4 years, 11 months ago

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No.4: Try right-click and open this image in new tab. I checked the problem 4 times and didn't see what is wrong.

alt text alt text

Samuraiwarm Tsunayoshi - 4 years, 11 months ago

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