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Re: BDMO 2016: National Junior/2

Capture.PNG Let $CF'G'$ be the reflection of $C'FG$. And as $CDE$ is the reflection of $C'DE$, we can say, $F'G'||DE||AB$ So, $(CF'G')=(C'FG)=80$. (ABC)=$\frac{1}{2}AB.CP=500$ or, $10CP=500$ or,$CP=50$ We know, $\triangle ABC~\triangle CF'G'$ $\frac {AB}{F'G'}=\frac {CP}{CQ}=k$ So, $F'G'=\frac{AB}{...
by Tasnood
Fri Dec 01, 2017 2:07 pm
 
Forum: Junior Level
Topic: BDMO 2016: National Junior/2
Replies: 4
Views: 452

Re: coxbazar divisional 2014 Q.no.-1

It can be solved in the way of Physics! The velocity of Kamrul $a$ m/s The velocity of Tusher $4a$ m/s The velocity of Avik $8a$ m/s After a limited time $t$, Kamrul, Tusher and Avik passed $at$ m, $4at$ m and $8at$ m respectively. Then, $8at-at=150$ $7at=105$ $13at=\frac{105*13}{7}$ As the total pa...
by Tasnood
Thu Nov 30, 2017 11:26 pm
 
Forum: Secondary Level
Topic: coxbazar divisional 2014 Q.no.-1
Replies: 1
Views: 402

Re: Sylhet - 2014

Let's find out the sum of some consecutive numbers greater than 976 with the lowest difference. That is:
$1+2+3+...+44=(41*44)/2=990$
990-976=14 ; the sum of the missing page number.

Look, $14=2+12=2+3+9=2+3+4+5$ ; that we want.
So, there were 4 pages missing.
[Maybe it is the easiest way :P ]
by Tasnood
Mon Nov 27, 2017 12:04 am
 
Forum: Secondary Level
Topic: Sylhet - 2014
Replies: 2
Views: 397

Re: BDMO Divisional_2014

You can't say EF||BC only to see that EF=BC/2. Because E and F mayn't be the midpoints. Thanks
by Tasnood
Thu Jan 26, 2017 10:08 pm
 
Forum: Divisional Math Olympiad
Topic: BDMO Divisional_2014
Replies: 2
Views: 272

Re: COMBINATORICS!!!

We assume that a number of 7 digits where at least 3 digits are same. As for example, if we take such a number, n = 111222 and the last digit will be 1 or 2. So, total combination: $\frac{7!}{3!4!}$ for the last digit 1 and same for 2. So, like (1,2), there are {(1,3),(1,4),(1,5),...,(1,9)} and 70 c...
by Tasnood
Thu Jan 26, 2017 8:47 pm
 
Forum: Junior Level
Topic: COMBINATORICS!!!
Replies: 4
Views: 336

Re: COMBINATORICS!!!

Not understand. In the second case, I see 7 has appeared once. So, where each digiit is appearing thrice (at least)?
by Tasnood
Tue Jan 24, 2017 10:47 pm
 
Forum: Junior Level
Topic: COMBINATORICS!!!
Replies: 4
Views: 336

BDMO Divisional_2014

The area of ABC and OBC triangle is 120 and 24 respectively. BC=16, EF=8. Find out the area of OEAF Quadrilateral.
by Tasnood
Tue Jan 24, 2017 10:42 pm
 
Forum: Divisional Math Olympiad
Topic: BDMO Divisional_2014
Replies: 2
Views: 272

Re: Chittagong Regional 2014

We assume that the 6-digit integer is: abcdef 11|abd and 11|cef where, c is not equal to d. Number of such 3-digit integer divisible by 11 is: (90-10+1)=81 If abd={110, 220, 330, 440, 550, 660, 770, 880, 990} where d=0, c must be different from d each time. For each of this abd, there are 81 choices...
by Tasnood
Sun Jan 22, 2017 8:11 pm
 
Forum: Divisional Math Olympiad
Topic: Chittagong Regional 2014
Replies: 1
Views: 280

Re: Dhaka '15 \10

Let BC and DF (Extended) meet at point Q. AE:ED=1:3, AE:AD=1:4, ED:AD=3:4 BF:AF=1:7, BF:AB=1:8, AF:AB=7:8 ∠FQB=∠DQC, BF||CD and ∠BFQ=∠CDQ. So, ΔBFQ~ΔCDQ, BQ/CQ = BF/CD = 1:8 ΔBFQ~ΔAFD, BF/AF = BQ/AD = 1/7 (BQ/AD) × (AD/ED) = 1/7 × 4/3 or, BQ/ED = 4/21 (BC/ED) + (BQ/ED) = 4/3 + 4/21 or, (BC+BQ)/ED = ...
by Tasnood
Sat Jan 21, 2017 5:33 pm
 
Forum: Junior Level
Topic: Dhaka '15 \10
Replies: 3
Views: 358

Re: Dhaka '15 \10

Capture.JPG Let DF and BC (Extended) meet at point Q. FB:AB=1:7, FB:CD=1:8. AE:ED=1:3, AD:ED=4:3. Between ΔDQC and ΔFQB, ∠Q is common and ∠BFQ=∠CDQ as AB||CD and QD is bisector. So, ΔDQC~ΔFQB. Then, FB:CD=QB:QC=1:8 ∠QFB=∠AFD, AD||BC||QC, ∠ADF=∠FQB, ΔAFD~ΔBFQ. QB:AD=BF:AF=1:7. QB/AD × AD/ED= 1/7 × 4...
by Tasnood
Fri Jan 20, 2017 8:46 am
 
Forum: Junior Level
Topic: Dhaka '15 \10
Replies: 3
Views: 358
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