×

# 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, 1 month ago

Sort by:

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. · 1 year, 1 month ago

Nicely done. · 1 year, 1 month 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. · 1 year, 1 month ago

So , finally you posted.I was waiting for you shraky :P · 1 year, 1 month ago

I had no internet for the last 2 days. I was transitioning to a more permanent wi-fi connection. · 1 year, 1 month ago