# 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
2 years, 9 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.

- 2 years, 9 months ago

Nicely done.

- 2 years, 9 months ago

@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.

- 2 years, 9 months ago

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

- 2 years, 9 months ago

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

- 2 years, 9 months ago

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.

- 1 year, 3 months ago

BRSC cyclic I mean

- 1 year, 3 months ago

- 1 year, 3 months ago

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.

- 1 year, 3 months ago

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

- 1 year, 3 months ago

@Sharky Kesa Is it from Australian School of Excellence?

- 1 year, 3 months ago

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

- 1 year, 3 months ago

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

- 2 years, 9 months ago