[OGC1] Online Geometry Camp: Day 1

Discussion on Bangladesh National Math Camp
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Samiun Fateeha Ira
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Re: [OGC1] Online Geometry Camp: Day 1

Unread post by Samiun Fateeha Ira » Mon Aug 26, 2013 6:26 am

Tusher Chakraborty wrote:Another Problem
(২)
AR, AD, BD, BE এবং CE পাঁচটি রেখাংশ। BE, AC ও AD কে যথাক্রমে P ও Q বিন্দুতে ছেদ করে। BD, AC ও EC কে R ও S বিন্দুতে ছেদ করে। EC, AD কে T বিন্দুতে ছেদ করে। যদি AP = AQ , $ \angle PAQ = 42^0, \angle ADB = x, \angle EBD = y $ ও $ \angle BRP=z $ হয় তবে $(y^2+xy+yz+zx)$ এর মান নির্ণয় কর।
AR, AD, BD, BE and CE are straight line segments. BE intersects AC and AD at P and Q respectively. BD intersects AC and EC at R and S respectively. EC intersects AD at T. If AP = AQ , $ \angle PAQ = 42^0, \angle ADB = x, \angle EBD = y $ and $ \angle BRP=z $ then what is the value of $(y^2+xy+yz+zx)$?

In ∆APQ , \PAQ= 42° & AP=AQ.
So, APQ= AQP= 69°
∴ BQD= 111°

In BQD, y=180-111-x = 69-x
In ARD, z= 42+ x

Now,
y^2+xy+yz+zx
= (69−x)^2+x(69−x)+(69−x)(42+x)+(42+x)x
= 4761- 138x+ x^2+69x−x^2+2898+69x−42x−x^2+42x+x^2
= 7659

Tahmid
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Re: [OGC1] Online Geometry Camp: Day 1

Unread post by Tahmid » Mon Aug 26, 2013 9:04 am

$\Delta APQ$ is isoscles and $\angle PAQ$=42.
so $\angle PAQ=\angle AQP$=(180-42)/2=69.

now compare for $\Delta BDQ$, x+y=$\angle BQA$=$\angle PQA$=$\angle AQP$=69.....1st result
and compare for $\Delta BRP$, y+z=180-$\angle BPR$=180-$\angle APQ$=180-69=111....2nd result

now multiply 1st and 2nd result.....(x+y) * (y+z)=69*111$\Leftrightarrow y^{2}+xy+yz+xz$=7659.

so ans=7659.

Neblina
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Re: [OGC1] Online Geometry Camp: Day 1

Unread post by Neblina » Mon Aug 26, 2013 10:51 am

Tusher Chakraborty wrote:Another Problem
(২)
AR, AD, BD, BE এবং CE পাঁচটি রেখাংশ। BE, AC ও AD কে যথাক্রমে P ও Q বিন্দুতে ছেদ করে। BD, AC ও EC কে R ও S বিন্দুতে ছেদ করে। EC, AD কে T বিন্দুতে ছেদ করে। যদি AP = AQ , $ \angle PAQ = 42^0, \angle ADB = x, \angle EBD = y $ ও $ \angle BRP=z $ হয় তবে $(y^2+xy+yz+zx)$ এর মান নির্ণয় কর।
AR, AD, BD, BE and CE are straight line segments. BE intersects AC and AD at P and Q respectively. BD intersects AC and EC at R and S respectively. EC intersects AD at T. If AP = AQ , $ \angle PAQ = 42^0, \angle ADB = x, \angle EBD = y $ and $ \angle BRP=z $ then what is the value of $(y^2+xy+yz+zx)$?
I think it should be AC not AR in "AR, AD, BD, BE এবং CE". I got confused while drawing it as in the next line BD intersects AC at R.

Tusher Chakraborty
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Re: [OGC1] Online Geometry Camp: Day 1

Unread post by Tusher Chakraborty » Mon Aug 26, 2013 12:52 pm

তাহমিদ এবং ইরা দুইজনেরই সমাধান ঠিক আছে :)
"Your present circumstances don't determine where you can go; they merely determine where you start." -Nido Qubein

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nayel
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Re: [OGC1] Online Geometry Camp: Day 1

Unread post by nayel » Mon Aug 26, 2013 1:00 pm

Samiun Fateeha Ira wrote:
nayel wrote:
Samiun Fateeha Ira wrote:এখন, ২য় ত্রিভুজ PQR এ উচ্চতা PS. এখানে PR=17, PS=DC=8 এবং SR=AD=15 হলে (PSR)=(ADC)=60 হয়।
এটা কি তুমি অনুমান করে পেয়েছ?

ঠিক অনুমান না। ত্রিভুজ ADC এর উচ্চতা এবং ভুমিকে ত্রিভুজ PSR এর ক্ষেত্রে যথাক্রমে ভুমি এবং উচ্চতা ধরে নিয়েছি।
ঠিক আছে।
"Everything should be made as simple as possible, but not simpler." - Albert Einstein

Ayantika rinti bose
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Re: [OGC1] Online Geometry Camp: Day 1

Unread post by Ayantika rinti bose » Wed Aug 28, 2013 12:53 am

Hello bhaia,

This is Ayantika Rinti Bose. Actually I was in the Lilaboyi camp till 24 august. And after coming home, I got severe cold and fever. That's why I couldn't join the online camp on 25th. But I wanted to do this barely. So now I am joining and giving my solution of question no 1. of day 1.

$ABC$ triangle's base is $16$. Divide it in $2$ same size by drawing a perpendicular from $A$. Name it $AD$. So $BD=CD=8$. By using Pythagoras' theorem,we find that $AD=15$.

Now make it double and the base of $PQR$ triangle. The other are the same means $PQ=PR=17$. And $QR=30$. The perpendicular distance of it is $8$. So its area is also $120$.

So another triangle is $PQR$ where $PQ=PR=17$ and $QR=30$.

By the way, how can I make a reply invisible?

Rinti.

Ayantika rinti bose
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Re: [OGC1] Online Geometry Camp: Day 1

Unread post by Ayantika rinti bose » Wed Aug 28, 2013 11:43 am

Solution of ques. no 2 of day 1:
In triangle $PQR$,$\angle PAQ=42$ and $AP=AQ$. Then,$\angle APQ=\angle AQP=69$.then $\angle EQT=\angle BPR=69$. So,in triangle $BQD$, $\angle BQD=111$. So $\angle QBD+\angle QDB=X+Y=69$.
In triangle $BPR$ $\angle BPR=69$. So, $\angle PBR+\angle BRP=Y+Z=111$
Now we have to find $y^2+xy+yz+zx=(x+y)(y+z)=69\times 111=7659$

brintodibyendu
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Re: [OGC1] Online Geometry Camp: Day 1

Unread post by brintodibyendu » Wed Aug 28, 2013 12:58 pm

first triangle is abc where ab~~ac~~17.bc~~16.if we devide bc then we get bd~~dc~~8.tgen using pythagoras we get ad~~15.then we get another triangle which hight is 8 and area is 120square metere.thus the nother triangle will get 30 17 17

Here ap~~aq.it means /_apq~~/_aqp~~180-42~~69.in triangle pbr y_|z~~180-69~~111.again in triangle bqd x_|y~~69.it means (x|_y)(y|_z)~~(111*69)~~7659

[Edited-]

first triangle is $ABC$ where $AB=AC=17.BC = 16$.if we divide (I guess it means bisect here) $BC$ then we get $BD =DC= 8$.Then using pythagoras we get $AD=15$ .then we get another triangle which height is $8$ and area is $120$ square meter.thus the another triangle will get $30, 17, 17$

Here $AP=AQ$.it means $\angle APQ = \angle AQP = \frac 12(180 - 42)^\circ = 69^\circ$.in $\triangle PBR$ $y + z = 180^\circ -69^\circ = 111^\circ$.again in $\triangle BQD$, $x+y = 69^\circ$ .it means $(x+y)(y+z) = 111*69 =7659$
Last edited by *Mahi* on Wed Aug 28, 2013 11:55 pm, edited 1 time in total.
Reason: Made it readble

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Phlembac Adib Hasan
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Re: [OGC1] Online Geometry Camp: Day 1

Unread post by Phlembac Adib Hasan » Wed Aug 28, 2013 4:56 pm

Ayantika rinti bose wrote:By the way, how can I make a reply invisible?

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I did it to your second post. Btw, help us using latex. It's not very hard. Whenever you write something mathematical, just write it between two \$ signs. For example, writing \$a^2+b^2=c^2\$ will become $a^2+b^2=c^2$. For details, see this article: How to use latex.

@brintodibyendu, বহু চেষ্টা করেও তোমার পোস্টের ঐ ~~ চিহ্ন দেওয়া কোডের মাথামুণ্ডু কিছুই বুঝতে পারলাম না। মডারেট(decipher) করা গেল না। দুঃখিত।
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Swargo
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Re: [OGC1] Online Geometry Camp: Day 1

Unread post by Swargo » Thu Sep 05, 2013 11:36 am

ধুর হালার কেউ আমাকে ফেসবুকে পর্যন্ত খবর দেয় নাই। ঐজন্যে এত দিন পর প্রথম দিনের সমস্যা দেখতে হচ্ছে।
প্রথম সমস্যাটার একটা বিকল্প সমাধান আমি জানি। এটা অনেকটা বইএর নিয়মে। নিচে দিলাম।

Solution to the prob no. 1:
Let '$b$' be the length of base. .

hence,

$\dfrac b 4\cdot \sqrt{1156-b^2}=120$

by generalizing the equation,we get

$1156-b^2=\dfrac {230400}{b^2}$

$\Longrightarrow b^4-1156b^2+230400=0$

$\Longrightarrow b^2(b^2-256)-900(b^2-256)=0$

by solving the equation,we get..

$b= 30, 16$

so $30,16$ are both the right ans. . . .

ANS: $17,17,30$

প্রব্লেম ২ এর সমাধান অন্যদের সাথে মিলে যাওয়ায় দিলাম না। খুবই কান্না পাচ্ছে, যে এত দিন পরে জয়েন করলাম।
Last edited by Phlembac Adib Hasan on Sun Sep 15, 2013 9:30 am, edited 1 time in total.
Reason: Latexed

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