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# Geometry (1st math Thailand POSN 2014)

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

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

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

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

• 4.1) If point $$I_{a}$$ is the center of excircle of $$\triangle ABC$$ which is opposite to angle $$A$$, prove that $$P$$ is the center of the circumcenter of $$\square ICI_{a}B$$

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

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

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

Note by Samuraiwarm Tsunayoshi
3 years, 2 months ago

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## Comments

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

- 3 years, 2 months ago

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Problem 1: Easy one. See here.

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

Problem 3: Consider $$\triangle ABC$$ where $$AD$$, $$BE$$, $$CF$$ are angle bisectors, with $$D,E,F$$ intersections with opposite sides. By Angle Bisector Theorem we have $$BD/DC=AB/AC$$, $$CE/EA=BC/AB$$, $$AF/FB=AC/BC$$. Multiplying these cancels Ceva. $$\square$$

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

Problem 5: Menelaus on $$\triangle QRS$$.

- 3 years, 1 month 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

- 3 years, 1 month ago

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2) $$\angle FDP=\angle ADF=\angle ACF$$ since $$ACDF$$ is cyclic. $$\angle ACF=\angle ECP=\angle FBP$$ since $$BCEF$$ is cyclic. Hence $$\angle FBP=\angle FDP \implies PDBF$$ is cyclic.

- 3 years, 2 months ago

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