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 $OH$ .

#Geometry #Sharky

Easy Math Editor

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Comments

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 - 5 years, 8 months ago

Nicely done.

Sharky Kesa - 5 years, 8 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.

Sharky Kesa - 5 years, 8 months ago

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

Nihar Mahajan - 5 years, 8 months ago

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

Sharky Kesa - 5 years, 8 months ago

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

Xuming Liang - 5 years, 8 months ago

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

Vishwash Kumar ΓΞΩ - 4 years, 3 months ago

@Sharky Kesa Is it from Australian School of Excellence?

Vishwash Kumar ΓΞΩ - 4 years, 3 months ago

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

Sharky Kesa - 4 years, 3 months ago

Vishwash Kumar ΓΞΩ - 4 years, 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.

Vishwash Kumar ΓΞΩ - 4 years, 3 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.

Alan Yan - 4 years, 3 months ago

BRSC cyclic I mean

Alan Yan - 4 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$