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.

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## Comments

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TopNewest1.) 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.

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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.

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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 \(\triangle QRS\).Log in to reply

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.

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