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

### 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

### 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

### 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.

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!)

That is a reply!

"Everything should be made as simple as possible, but not simpler." - Albert Einstein

### 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, $AE\times34=17\times16=272.\therefore AE=8.\therefore EC=26$.

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"

### 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 $Power$ $of$ $point$,$AP \times 2\sqrt{5}=4$.$\therefore$ $AP=\frac{2} {\sqrt{5}}$.$\therefore PQ=\frac{8} {\sqrt{5}}$.

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"

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

$\text{Another solution of problem 6}$:

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 $\Delta AOC,\angle OAC=\angle PAC=\angle PKC=\angle CKL=\angle DKN,\angle ACB=\angle ABC=\angle BNK= \angle DNK$.$\therefore$ $\angle KDN=\angle AOC=90^{\circ}$.$\therefore$ $AP\perp BC$.

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 $\Delta AOC,\angle OAC=\angle PAC=\angle PKC=\angle CKL=\angle DKN,\angle ACB=\angle ABC=\angle BNK= \angle DNK$.$\therefore$ $\angle KDN=\angle AOC=90^{\circ}$.$\therefore$ $AP\perp BC$.

"Questions we can't answer are far better than answers we can't question"