# 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
5 years ago

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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.

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.

- 5 years ago

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 $\triangle QRS$.

- 4 years, 11 months ago

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

- 4 years, 11 months ago