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Xuming's Synthetic Geometry Group - Sharky's Submissions

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 \(OH\).

Note by Sharky Kesa
1 year, 8 months ago

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Question 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, 8 months ago

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@Alan Yan Nicely done. Sharky Kesa · 1 year, 8 months ago

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@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, 8 months ago

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@Sharky Kesa So , finally you posted.I was waiting for you shraky :P Nihar Mahajan · 1 year, 8 months ago

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@Nihar Mahajan I had no internet for the last 2 days. I was transitioning to a more permanent wi-fi connection. Sharky Kesa · 1 year, 8 months ago

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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 · 2 months, 4 weeks ago

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@Alan Yan BRSC cyclic I mean Alan Yan · 2 months, 4 weeks ago

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Rohit Camfar · 2 months, 4 weeks ago

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@Rohit Camfar Solution: 2 :: (\Since F lies on Circumcircle of ADE. It is sufficient enough to prove that M'F = M'N.

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 · 2 months, 4 weeks ago

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Here is an another one for the second without using Nine Point Circles. Rohit Camfar · 2 months, 4 weeks ago

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@Rohit Camfar @Sharky Kesa Is it from Australian School of Excellence? Rohit Camfar · 2 months, 4 weeks ago

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@Rohit Camfar Yes, this was from the Junior problems in the School of Excellence from 2 years ago. Sharky Kesa · 2 months, 4 weeks ago

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Hints: 1. power of a point/radical axis configuration/similarity 2. reflect a certain point. Xuming Liang · 1 year, 8 months ago

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