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 ABCABC has circumcircle Γ\Gamma. The tangent at AA to Γ\Gamma intersects BCBC at PP. Let MM be the midpoint of the segment APAP. Let RR be the second intersection point of BMBM with Γ\Gamma. Let SS be the second intersection point of PRPR with Γ\Gamma. Prove CSCS is parallel to PAPA.

2. Let OO be the circumcentre of acute ΔABC\Delta ABC, HH be the orthocentre. Let ADAD be the altitude of ΔABC\Delta ABC from AA, and let the perpendicular bisector of AOAO intersect BCBC at EE. Prove that the circumcircle of ΔADE\Delta ADE passes through the midpoint of OHOH.

Note by Sharky Kesa
4 years ago

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Question 2. Let's call the midpoint of AOAO and OHOH, PP and NN respectively. Since APE=ADE\angle APE = \angle ADE , this implies that quadrilateral APDEAPDE is cyclic. Therefore it suffices to prove that quadrilateral APNDAPND is cyclic as well.

Since OPAO=ONOH\frac{OP}{AO} = \frac{ON}{OH} , we know that PNADPN || AD.

We also know that NN is the center of the nine-point circle, and DD is on the circle, therefore ND=R2ND = \frac{R}{2} . We also know that PA=AO2=R2PA = \frac{AO}{2} = \frac{R}{2} . This implies that APNDAPND is a isosceles trapezoid which implies that this is cyclic and we are done.

Alan Yan - 4 years ago

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Nicely done.

Sharky Kesa - 4 years 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 - 4 years ago

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So , finally you posted.I was waiting for you shraky :P

Nihar Mahajan - 4 years ago

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

Sharky Kesa - 4 years ago

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Hints: 1. power of a point/radical axis configuration/similarity 2. reflect a certain point.

Xuming Liang - 4 years ago

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Here is an another one for the second without using Nine Point Circles.

The almighty knows it all. - 2 years, 7 months ago

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@Sharky Kesa Is it from Australian School of Excellence?

The almighty knows it all. - 2 years, 7 months ago

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

Sharky Kesa - 2 years, 7 months ago

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The almighty knows it all. - 2 years, 7 months ago

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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.

The almighty knows it all. - 2 years, 7 months ago

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Question 1:

From Power of a Point,

MB×MR=MA2=MB2MB \times MR = MA^2 = MB^2

Thus, MPBMRP\triangle MPB \sim \triangle MRP. Thus, APC=PRB=SCP \angle APC = \angle PRB = \angle SCP

where the last follows from SCQR cyclic.

Thus we can conclude.

Alan Yan - 2 years, 7 months ago

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BRSC cyclic I mean

Alan Yan - 2 years, 7 months ago

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