## [OGC1] Online Geometry Camp: Day 6 (EXAM!)

Discussion on Bangladesh National Math Camp
nayel
Posts: 268
Joined: Tue Dec 07, 2010 7:38 pm
Location: Dhaka, Bangladesh or Cambridge, UK

### Re: [OGC1] Online Geometry Camp: Day 6 (EXAM!)

To those looking for Bengali translations of problems: for goodness' sake please start practising maths in English if you are aiming for the IMO! You won't get questions in Bengali there!
"Everything should be made as simple as possible, but not simpler." - Albert Einstein

nayel
Posts: 268
Joined: Tue Dec 07, 2010 7:38 pm
Location: Dhaka, Bangladesh or Cambridge, UK

### Re: [OGC1] Online Geometry Camp: Day 6 (EXAM!)

I've replied to (hopefully) everyone who has sent me solutions. If you didn't get a reply, let me know asap.
"Everything should be made as simple as possible, but not simpler." - Albert Einstein

Siam H.
Posts: 4
Joined: Wed Aug 28, 2013 10:46 pm
Location: Dhaka

### Re: [OGC1] Online Geometry Camp: Day 6 (EXAM!)

I didn't get a reply from you. I however got two mails where you said you received my solutions.

My email ID is hsiam261@gmail.com.

I have sent you images. My handwriting isn't that good. So I apologize. Please zoom in to read.

nayel
Posts: 268
Joined: Tue Dec 07, 2010 7:38 pm
Location: Dhaka, Bangladesh or Cambridge, UK

### Re: [OGC1] Online Geometry Camp: Day 6 (EXAM!)

That is a reply!
"Everything should be made as simple as possible, but not simpler." - Albert Einstein

tanmoy
Posts: 289
Joined: Fri Oct 18, 2013 11:56 pm

### Re: [OGC1] Online Geometry Camp: Day 6 (EXAM!)

$Solution$ $of$ $1$:
Join $C,D$.By Pythagorus's theorem,$CD=30,AC=34$.$\therefore$ By Power of point, $AE\times34=17\times16=272.\therefore AE=8.\therefore EC=26$.
"Questions we can't answer are far better than answers we can't question"

tanmoy
Posts: 289
Joined: Fri Oct 18, 2013 11:56 pm

### Re: [OGC1] Online Geometry Camp: Day 6 (EXAM!)

$Solution$ $of$ $2$:
Draw a tangent at point $Q$.Suppose,it meets $AB$ at point $R$ and $CD$ at $S$.Line $RS$ is perpendicular to $AB$ and $CD$.Now,$AR=4$ and $RQ=2$.$\therefore AQ=2\sqrt{5}$.Suppose,the inscribed circle touches $AB$ at $F$.Then by $Power$ $of$ $point$,$AP \times 2\sqrt{5}=4$.$\therefore$ $AP=\frac{2} {\sqrt{5}}$.$\therefore PQ=\frac{8} {\sqrt{5}}$.
"Questions we can't answer are far better than answers we can't question"

tanmoy
Posts: 289
Joined: Fri Oct 18, 2013 11:56 pm
$\text{Another solution of problem 6}$:
Suppose,the line through $K$ parallel to $BC$ intersects $AB$ at $N$ and meets the circumcircle at $L$.Let $AP\cap BC=O$ and $CK\cap AB=D$.Now,in $\Delta DKN$ and $\Delta AOC,\angle OAC=\angle PAC=\angle PKC=\angle CKL=\angle DKN,\angle ACB=\angle ABC=\angle BNK= \angle DNK$.$\therefore$ $\angle KDN=\angle AOC=90^{\circ}$.$\therefore$ $AP\perp BC$.