My problems are from the Australian School of Excellence I went to last year from the worksheets.

**1.** Acute triangle \(ABC\) has circumcircle \(\Gamma\). The tangent at \(A\) to \(\Gamma\) intersects \(BC\) at \(P\). Let \(M\) be the midpoint of the segment \(AP\). Let \(R\) be the second intersection point of \(BM\) with \(\Gamma\). Let \(S\) be the second intersection point of \(PR\) with \(\Gamma\). Prove \(CS\) is parallel to \(PA\).

**2.** Let \(O\) be the circumcentre of acute \(\Delta ABC\), \(H\) be the orthocentre. Let \(AD\) be the altitude of \(\Delta ABC\) from \(A\), and let the perpendicular bisector of \(AO\) intersect \(BC\) at \(E\). Prove that the circumcircle of \(\Delta ADE\) passes through the midpoint of \(OH\).

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TopNewestQuestion 2. Let's call the midpoint of \(AO\) and \(OH\), \(P\) and \(N\) respectively. Since \(\angle APE = \angle ADE \), this implies that quadrilateral \(APDE\) is cyclic. Therefore it suffices to prove that quadrilateral \(APND\) is cyclic as well.

Since \(\frac{OP}{AO} = \frac{ON}{OH} \), we know that \(PN || AD\).

We also know that \(N\) is the center of the nine-point circle, and \(D\) is on the circle, therefore \(ND = \frac{R}{2} \). We also know that \(PA = \frac{AO}{2} = \frac{R}{2} \). This implies that \(APND\) is a isosceles trapezoid which implies that this is cyclic and we are done. – Alan Yan · 1 year, 12 months ago

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– Sharky Kesa · 1 year, 11 months ago

Nicely done.Log in to reply

@Nihar Mahajan ,@Mehul Arora , @Surya Prakash , @Agnishom Chattopadhyay , @Alan Yan , @Shivam Jadhav , @Swapnil Das , @Mardokay Mosazghi , @Kushagra Sahni , @Xuming Liang Post your solutions please. Sorry for the late submission. – Sharky Kesa · 1 year, 12 months ago

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– Nihar Mahajan · 1 year, 12 months ago

So , finally you posted.I was waiting for you shraky :PLog in to reply

– Sharky Kesa · 1 year, 12 months ago

I had no internet for the last 2 days. I was transitioning to a more permanent wi-fi connection.Log in to reply

Question 1:

From Power of a Point,

\[MB \times MR = MA^2 = MB^2\]

Thus, \(\triangle MPB \sim \triangle MRP\). Thus, \[ \angle APC = \angle PRB = \angle SCP\]

where the last follows from SCQR cyclic.

Thus we can conclude. – Alan Yan · 6 months, 3 weeks ago

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– Alan Yan · 6 months, 3 weeks ago

BRSC cyclic I meanLog in to reply

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OM' || FN. Also, OM = 1/2 AH (Euler rule) And FN = 1/2 AH (Midpoint theorem) => OM = FN.

=> MNFO is a ||gm. => NM || AF and

NM= FO =AF'.' NM || AF => ∠ AFP = ∠ MNP , ∠ APF = ∠ MPN and ON = AF. => Tri. AFP = Tri. MNP.

=> AP = PM and

FP = NP.---- (1)Also AP = PM => M'P || EC =>

M'P perp. FN---- (2)The two results (1) and (2) tells us that M'F = M'N/)

Sorry for the inconvenience of Latex. I have just learnt to write italics only. – Rohit Camfar · 6 months, 3 weeks ago

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Here is an another one for the second without using Nine Point Circles. – Rohit Camfar · 6 months, 3 weeks ago

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@Sharky Kesa Is it from Australian School of Excellence? – Rohit Camfar · 6 months, 3 weeks ago

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– Sharky Kesa · 6 months, 3 weeks ago

Yes, this was from the Junior problems in the School of Excellence from 2 years ago.Log in to reply

Hints: 1. power of a point/radical axis configuration/similarity 2. reflect a certain point. – Xuming Liang · 1 year, 12 months ago

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