These problems are my first submissions to Xuming's Geometry group.

However they are for the purpose of sharing the Geometry problems from Pakistan's First Round (for 2016)

Q1.The tangents at $A,B$ to the circumcircle$\omega$ of Triangle $ABC$ meet at $T$. The line through $T||AC$ meets $BC$ at $D$. Prove that $AD=CD$.

Q2. In a right angled triangle $ABC, \hat{C}=90^{\circ}, CD\perp AB$ at $D$ and the angle bisector of $\hat{B}$ intersects $CD,AC$ at $O,E$ respectively. Through $O$ introduce $FG \parallel AB$ such that $FG$ intersects $AC,BC$ at $F,G$ respectively. Prove that $AF=CE$.

No vote yet

1 vote

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in`\(`

...`\)`

or`\[`

...`\]`

to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestDue to lack of time I have uploaded pics of my solution to problem 1.

Log in to reply

Good job Nihar ! The crux move was to identify that TBDA is cyclic, by wishful thinking.

Log in to reply

absolutely!

Log in to reply

@Calvin Lin @Xuming Liang Here is my submission!

@Nihar Mahajan ,@Sharky Kesa I thought you would be interested!

Log in to reply

Ill add in the solutions when requested. (Not before a week though)

Log in to reply

1st one is just some simple angle chase using alternate segment theorem as proved in the solution by @Nihar Mahajan.

2nd one is also just trivial: Solution to II.

$\Delta DBC \sim \Delta CBA$

=> $\dfrac{DB}{CB} = \dfrac{CB}{AB}$

=> $\dfrac{DO}{OC} = \dfrac{CE}{AE}$ --- (Int. Angle Bisector Theorem)

=> $\dfrac{AF}{CF} = \dfrac{CE}{AE}$

=> $\dfrac{AF}{AC} = \dfrac{CE}{AC}$ ---- (By rule of Dividendo)

=> $AF = AC$

K.I.P.K.I.G

Log in to reply