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

Posted: Sat Aug 31, 2013 5:07 pm
by nayel
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!

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

Posted: Sat Aug 31, 2013 5:57 pm
by nayel
I've replied to (hopefully) everyone who has sent me solutions. If you didn't get a reply, let me know asap.

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

Posted: Sat Aug 31, 2013 7:33 pm
by Siam H.
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.

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

Posted: Sat Aug 31, 2013 9:21 pm
by nayel
That is a reply!

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

Posted: Sat Jan 03, 2015 6:36 pm
by tanmoy
$Solution$ $of$ $1$:
Join $C,D$.By Pythagorus's theorem,$CD=30,AC=34$.$\therefore$ By Power of point, $AE\times34=17\times16=272.\therefore AE=8.\therefore EC=26$. :)

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

Posted: Sat Jan 03, 2015 7:38 pm
by tanmoy
$Solution$ $of$ $2$:
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 $Power$ $of$ $point$,$AP \times 2\sqrt{5}=4$.$\therefore$ $AP=\frac{2} {\sqrt{5}}$.$\therefore PQ=\frac{8} {\sqrt{5}}$. :)

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

Posted: Mon Jan 05, 2015 4:41 pm
by tanmoy
$\text{Another solution of problem 6}$:
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 $\Delta AOC,\angle OAC=\angle PAC=\angle PKC=\angle CKL=\angle DKN,\angle ACB=\angle ABC=\angle BNK= \angle DNK$.$\therefore$ $\angle KDN=\angle AOC=90^{\circ}$.$\therefore$ $AP\perp BC$. :)