[OGC1] Online Geometry Camp: Day 4

Discussion on Bangladesh National Math Camp
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nayel
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[OGC1] Online Geometry Camp: Day 4

Unread post by nayel » Wed Aug 28, 2013 10:14 am

Post all day 4 discussion threads links here.
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Tusher Chakraborty
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Re: [OGC1] Online Geometry Camp: Day 4

Unread post by Tusher Chakraborty » Wed Aug 28, 2013 12:30 pm

আজকে ক্যাম্পে আলোচনার শেষ দিন। কালকে Day-1 ও Day-2 এর ওপর পরীক্ষা। সকাল ৮ টায় প্রশ্ন আপলোড দেওয়া হবে।
আজকের টপিকঃ
বৃত্ত সংক্রান্ত উপপাদ্য- ৪.১ হতে ৪.৮(মাধ্যমিক উচ্চতর জ্যামিতি)
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Tusher Chakraborty
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Re: [OGC1] Online Geometry Camp: Day 4

Unread post by Tusher Chakraborty » Wed Aug 28, 2013 3:38 pm

একদম সহজ একটা প্রবলেম দিয়ে শুরু করি। এটা এবারের ঢাকা-১ এ ছিল।
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"Your present circumstances don't determine where you can go; they merely determine where you start." -Nido Qubein

Nahraf
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Re: [OGC1] Online Geometry Camp: Day 4

Unread post by Nahraf » Wed Aug 28, 2013 3:59 pm

ভাইয়া, এটাচমেন্টটা শো করছে না৷ একটু লেখে দিবেন? অথবা pdf?
"And that there is not for man except that for which he strives."
Al Quran, 53:39

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Mursalin
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Re: [OGC1] Online Geometry Camp: Day 4

Unread post by Mursalin » Wed Aug 28, 2013 4:01 pm

Here are a couple of problems for you to practise. These problems are from older sets in Brilliant.org.

1)A circle $\Gamma$ cuts the sides of a equilateral triangle $ABC$ at $6$ distinct points. Specifically, $\Gamma$ intersects $AB$ at points $D$ and $E$ such that $A$,$D$,$E$,$B$ lie in order. $\Gamma$ intersects $BC$ at points $F$ and $G$ such that $B$,$F$,$G$,$C$ lie in order. $\Gamma$ intersects $CA$ at points $H$ and $I$ such that $C$,$H$,$I$,$A$ lie in order. If $|AD|=3$, $|DE|=39$, $|EB|=6$ and $|FG|=21$, what is the value of $|HI|^2$?

Hint: Try using the intersecting secants theorem. See the original problem here.

2)Circles $\Gamma_1$ and $\Gamma_2$ have centers $X$ and $Y$ respectively. They intersect at points $A$ and $B$, such that angle $XAY$ is obtuse. The line $AX$ intersects $\Gamma_2$ again at $P$, and the line $AY$ intersects $\Gamma_1$ again at $Q$. Lines $PQ$ and $XY$ intersect at $G$, such that $Q$ lies on line segment $GP$. If $GQ=255$, $GP=266$ and $GX=190$, what is the length of $XY$?

Hint: Power-of-a-point or noticing cyclic quads could help. The original problem is here.

3)$ABC$ is a triangle with $AC=139$ and $BC=178$. Points $D$ and $E$ are the midpoints of $BC$ and $AC$ respectively. Given that $AD$ and $BE$ are perpendicular to each other, what is the length of $AB$?

Hint: The centroid of a triangle trisects its medians. The original problem.

4)$ABCD$ is a trapezoid(US) or Trapezium(UK) with parallel sides $AB$ and $CD$. $\Gamma$ is an inscribed circle of $ABCD$, and tangential to sides $AB$,$BC$,$CD$ and $AD$ at the points $E$,$F$,$G$ and $H$ respectively. If $AE=2$,$BE=3$, and the radius of $\Gamma$ is $12$, what is the length of $CD$?

Hint: There are multiple ways to solve this problem. Are you familiar with Pitot's Theorem? See the original problem here.

Have fun solving these problems!
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nowshin
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Re: [OGC1] Online Geometry Camp: Day 4

Unread post by nowshin » Wed Aug 28, 2013 4:19 pm

বৃত্তদ্বয়ের মাঝে যে ছায়া আছে তার ক্ষেত্রফল $\dfrac{\pi}{2} -1$ এবং উপরের ছায়াযুক্ত অংশের ক্ষেত্রফলও $\dfrac{\pi}{2} -1$. অতএব এদের পার্থক্য $0$. সুতরাং $a=0$. :roll:

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Re: [OGC1] Online Geometry Camp: Day 4

Unread post by Tusher Chakraborty » Wed Aug 28, 2013 4:23 pm

nowshin wrote:বৃত্তদ্বয়ের মাঝে যে ছায়া আছে তার ক্ষেত্রফল $\dfrac{\pi}{2} -1$ এবং উপরের ছায়াযুক্ত অংশের ক্ষেত্রফলও $\dfrac{\pi}{2} -1$. অতএব এদের পার্থক্য $0$. সুতরাং $a=0$. :roll:
Details please :)
"Your present circumstances don't determine where you can go; they merely determine where you start." -Nido Qubein

Tusher Chakraborty
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Re: [OGC1] Online Geometry Camp: Day 4

Unread post by Tusher Chakraborty » Wed Aug 28, 2013 4:26 pm

Nahraf wrote:ভাইয়া, এটাচমেন্টটা শো করছে না৷ একটু লেখে দিবেন? অথবা pdf?
Image save করে নিয়ে দেখো।
"Your present circumstances don't determine where you can go; they merely determine where you start." -Nido Qubein

Neblina
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Re: [OGC1] Online Geometry Camp: Day 4

Unread post by Neblina » Wed Aug 28, 2013 4:58 pm

Area of large circle is $4\pi$
Let the area of intersection between the small circles be x
Area of one small circle is pi.
So total area of four small circles= $$4\pi-4x$$.
Let area of upper shadow part be y
So $$4\pi-(4\pi-4x)=4y$$
$$4\pi-4\pi+4x=4y$$
This gives up x=y
So the difference between the two shaded area is $0$. so $a=0$

Tusher Chakraborty
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Re: [OGC1] Online Geometry Camp: Day 4

Unread post by Tusher Chakraborty » Wed Aug 28, 2013 5:09 pm

@Neblina, perfect :)
"Your present circumstances don't determine where you can go; they merely determine where you start." -Nido Qubein

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