[OGC1] Online Geometry Camp: Day 2
[OGC1] Online Geometry Camp: Day 2
Post all the day 2 discussion threads links here. (Sorry for the delay, there was a power cut and so no internet.)
"Everything should be made as simple as possible, but not simpler."  Albert Einstein

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Re: [OGC1] Online Geometry Camp: Day 2
আজকের টপিক হলঃ
উপপাদ্য২.১ হতে ২.৩ ও ৩.১ হতে ৩.১০(মাধ্যমিক উচ্চতর জ্যামিতি)।
Ceva’s Theorem and application.
https://www.youtube.com/channel/UCVSITM ... hOESNsitCA
এই লিঙ্কে Ceva's Theorem ও এর application নিয়ে দুটি ভিডিও দেওয়া আছে। দেখে নিতে পারো।
বিকালে আবার প্রবলেম নিয়ে আলোচনা হবে।
উপপাদ্য২.১ হতে ২.৩ ও ৩.১ হতে ৩.১০(মাধ্যমিক উচ্চতর জ্যামিতি)।
Ceva’s Theorem and application.
https://www.youtube.com/channel/UCVSITM ... hOESNsitCA
এই লিঙ্কে Ceva's Theorem ও এর application নিয়ে দুটি ভিডিও দেওয়া আছে। দেখে নিতে পারো।
বিকালে আবার প্রবলেম নিয়ে আলোচনা হবে।
"Your present circumstances don't determine where you can go; they merely determine where you start." Nido Qubein
Re: [OGC1] Online Geometry Camp: Day 2
প্রথম problemটা আমিই শুরু করলাম।
$\triangle ABC$ এ $AB=\sqrt{41}$, $AC=\sqrt{164}$. $\angle BAC$ এর internal সমদ্বিখণ্ডক $BC$ কে $D$ বিন্দুতে ছেদ করে। যদি $AD$ আর $DC$ positive integer হয়, তাহলে $AD+DC$ এর মান কত?
In triangle $ABC$, $AB=\sqrt{41}$ and $AC=\sqrt{164}$ . The internal angle bisector of $BAC$ meets $BC$ at $D$. Given that lengths $AD$ and $DC$ are positive integers, what is the value of $AD+DC$?
Hint: Angle bisector theorem, cosine law. You could even use a direct formula for the internal angle bisector of a triangle (which can be proved using Stewart's Theorem, which could be proved by the cosine law]
Source: An old problem set from brilliant.org. You can see the original problem here https://brilliant.org/mathematicsprobl ... 0oh1NWooyK.
Happy problem solving!
$\triangle ABC$ এ $AB=\sqrt{41}$, $AC=\sqrt{164}$. $\angle BAC$ এর internal সমদ্বিখণ্ডক $BC$ কে $D$ বিন্দুতে ছেদ করে। যদি $AD$ আর $DC$ positive integer হয়, তাহলে $AD+DC$ এর মান কত?
In triangle $ABC$, $AB=\sqrt{41}$ and $AC=\sqrt{164}$ . The internal angle bisector of $BAC$ meets $BC$ at $D$. Given that lengths $AD$ and $DC$ are positive integers, what is the value of $AD+DC$?
Hint: Angle bisector theorem, cosine law. You could even use a direct formula for the internal angle bisector of a triangle (which can be proved using Stewart's Theorem, which could be proved by the cosine law]
Source: An old problem set from brilliant.org. You can see the original problem here https://brilliant.org/mathematicsprobl ... 0oh1NWooyK.
Happy problem solving!
Re: [OGC1] Online Geometry Camp: Day 2
As AD is the angle bisector $$\frac{BD}{DC}$$= $$\sqrt{\frac{41}{164}}$$= $$\frac{1}{2}$$. So BD=$$\frac{DC}{2}$$.
Using cosine rule,
$$AC^2=AD^2+DC^22AD.DC.cosx$$.....1st equation
$$AB^2=AD^2+\frac{DC^2}{4}AD.DC.cos(180x)$$
As cos(180x)=cosx,
so $$AB^2=AD^2+\frac{DC^2}{4}+AD.DC.cosx$$....2nd equation
$$AB^2$$= 164 and $$AC^2$$=41
multiplying the 2nd equation by 2 and adding it with equation 1, we get
$$246=3AD^2+\frac{3}{2}DC^2$$
dividing by 3 in both sides
$$AD^2+\frac{DC^2}{2}=82$$
Now i got stuck in this part, but as the question says that AD and DC are positive integers, i started taking values of AD,like, AD=1, AD=2......and found that when AD=8 DC=6,and both AD and DC are intergers!
So AD+DC= 14!
Using cosine rule,
$$AC^2=AD^2+DC^22AD.DC.cosx$$.....1st equation
$$AB^2=AD^2+\frac{DC^2}{4}AD.DC.cos(180x)$$
As cos(180x)=cosx,
so $$AB^2=AD^2+\frac{DC^2}{4}+AD.DC.cosx$$....2nd equation
$$AB^2$$= 164 and $$AC^2$$=41
multiplying the 2nd equation by 2 and adding it with equation 1, we get
$$246=3AD^2+\frac{3}{2}DC^2$$
dividing by 3 in both sides
$$AD^2+\frac{DC^2}{2}=82$$
Now i got stuck in this part, but as the question says that AD and DC are positive integers, i started taking values of AD,like, AD=1, AD=2......and found that when AD=8 DC=6,and both AD and DC are intergers!
So AD+DC= 14!
 Samiun Fateeha Ira
 Posts: 23
 Joined: Sat Aug 24, 2013 7:08 pm
 Location: Dhaka, Bangladesh
Re: [OGC1] Online Geometry Camp: Day 2
$AD$ bisects $\angle BAC$. So, $\dfrac {AB}{BC}= \dfrac{BD}{DC}$
$\sqrt {\dfrac {41}{164}} = \dfrac {BD}{DC}\Longrightarrow \dfrac 1 2=\dfrac{BD}{DC}\Longrightarrow BD= \dfrac 1 2 DC$.
From Stewart's theorem, in triangle ABC,
\begin{align*} & AC^2\cdot BD+AB^2\cdot DC= BC(AD^2+ BD\cdot DC)\\
& \frac {164}{2}\cdot DC + 41DC= \frac 3 2 \cdot DC(AD^2+ DC^2/2)\\
& AD^2+ \frac {DC^2}{2}= 82\Longrightarrow 2AD^2+ DC^2= 164\end{align*}
Here, after trying some values, I got $AD=8$ & $DC=6$. So, $AD+DC=14$.
$\sqrt {\dfrac {41}{164}} = \dfrac {BD}{DC}\Longrightarrow \dfrac 1 2=\dfrac{BD}{DC}\Longrightarrow BD= \dfrac 1 2 DC$.
From Stewart's theorem, in triangle ABC,
\begin{align*} & AC^2\cdot BD+AB^2\cdot DC= BC(AD^2+ BD\cdot DC)\\
& \frac {164}{2}\cdot DC + 41DC= \frac 3 2 \cdot DC(AD^2+ DC^2/2)\\
& AD^2+ \frac {DC^2}{2}= 82\Longrightarrow 2AD^2+ DC^2= 164\end{align*}
Here, after trying some values, I got $AD=8$ & $DC=6$. So, $AD+DC=14$.
Last edited by Phlembac Adib Hasan on Tue Aug 27, 2013 1:33 am, edited 1 time in total.
Reason: Latexed
Reason: Latexed
Re: [OGC1] Online Geometry Camp: Day 2
@Neblina @ Ira
Both of you are right!
But there is room for improvement. Instead of trying out values for $AD$ and $DC$, take a look below.
Both of you have reached this equation:
$AD^2$+$\frac{DC^2}{2}=82$.
Now you could use parity to conclude that both $AD$ and $DC$ are even numbers. So that cuts off some of the workload. You could also bound $AD$ [the maximum value of $AD$ is 9]. That way, you could stop plugging in values at a certain point as opposed to doing it indefinitely.
If I rephrased the problem like this: "If $AD \neq 8$, prove that both $AD$ and $DC$ can not be integers.", how would you approach? Would you keep on trying values forever? My point exactly!
You have both arrived at the correct numerical answer, but you haven't proved that it is the only answer.
Hope this helps!
PS: @admin. I'm unable to see any sort of Bangla text. Is there something wrong with me?
Both of you are right!
But there is room for improvement. Instead of trying out values for $AD$ and $DC$, take a look below.
Both of you have reached this equation:
$AD^2$+$\frac{DC^2}{2}=82$.
Now you could use parity to conclude that both $AD$ and $DC$ are even numbers. So that cuts off some of the workload. You could also bound $AD$ [the maximum value of $AD$ is 9]. That way, you could stop plugging in values at a certain point as opposed to doing it indefinitely.
If I rephrased the problem like this: "If $AD \neq 8$, prove that both $AD$ and $DC$ can not be integers.", how would you approach? Would you keep on trying values forever? My point exactly!
You have both arrived at the correct numerical answer, but you haven't proved that it is the only answer.
Hope this helps!
PS: @admin. I'm unable to see any sort of Bangla text. Is there something wrong with me?
"Be boring. It's the only way get work done."
 Austin Kleon
 Austin Kleon
Re: [OGC1] Online Geometry Camp: Day 2
Oww yah...didn't think of that, but would've if the question had asked for "values" of AD+DC. But hadn't thought about the fact that AD could have a maximum value.So...thanks.
Re: [OGC1] Online Geometry Camp: Day 2
I have also solved the problem by using stewart's theorem like Samiun Fateeha Ira and i also unable to see any bangla text ....

 Posts: 27
 Joined: Wed Oct 24, 2012 10:16 pm
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Re: [OGC1] Online Geometry Camp: Day 2
আজকের দিনের দ্বিতীয় প্রব্লেমঃ
(২)
$ABCD$ আয়তাকার কাগজের টুকরো যেখানে $AB=5$, $BC=12$। কাগজটিকে এমনভাবে ভাঁজ করা হল যেন $A$ বিন্দুটি $C$ এর উপর পড়ে। কাগজটি এতে $PQ$ রেখাংশ দ্বারা দুই ভাগে বিভক্ত হয়। $PQ$ এর দৈর্ঘ্য কত?
$ABCD$ is a rectangular paper where $AB=5$, $BC=12$. The paper was folded in such a way so that the point $A$ meets the point $C$. As a result the paper becomes divided into two regions by a line segment $PQ$. Find the length of $PQ$.
(২)
$ABCD$ আয়তাকার কাগজের টুকরো যেখানে $AB=5$, $BC=12$। কাগজটিকে এমনভাবে ভাঁজ করা হল যেন $A$ বিন্দুটি $C$ এর উপর পড়ে। কাগজটি এতে $PQ$ রেখাংশ দ্বারা দুই ভাগে বিভক্ত হয়। $PQ$ এর দৈর্ঘ্য কত?
$ABCD$ is a rectangular paper where $AB=5$, $BC=12$. The paper was folded in such a way so that the point $A$ meets the point $C$. As a result the paper becomes divided into two regions by a line segment $PQ$. Find the length of $PQ$.
"Your present circumstances don't determine where you can go; they merely determine where you start." Nido Qubein
Re: [OGC1] Online Geometry Camp: Day 2
For some weird reason I can't read the Bangla texts either! This has never happened with me before. I can however read it after pressing the 'quote' button.
"Everything should be made as simple as possible, but not simpler."  Albert Einstein