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BdMO Online Forum • View topic - Dhaka '15 \10

Dhaka '15 \10

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Dhaka '15 \10

Post Number:#1  Unread postby Absur Khan Siam » Thu Jan 19, 2017 12:16 pm

$ABCD$ is a parallelogram. $E$ intersects $AD$ as $AE:ED =1:3$ and $F$ intersects $AB$ as $AF:FB=7:1$. $CE$ and $DF$ meets at point $P$. $CP:PE =$?
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid
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Re: Dhaka '15 \10

Post Number:#2  Unread postby Tasnood » Fri Jan 20, 2017 8:46 am

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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 × 4/3
or, QB/ED = 4/21
Again, AD:ED=BC:ED=4:3
So, QB/ED + BC/ED = 4/21 + 4/3
or, (QB+BC)/ED = (4+4×7)/21
or, QC/ED = 32/21,
We know, ΔQPC~EPD as AD||BC||QC||ED.
So, CP:PE = QC:ED = 32/21.
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Re: Dhaka '15 \10

Post Number:#3  Unread postby Tasnood » Sat Jan 21, 2017 5:33 pm

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 = (28+4)/21
or, CQ/ED = 32/21

As ΔCPQ~ΔEPD, CP:PE = CQ:ED = 32:21
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Re: Dhaka '15 \10

Post Number:#4  Unread postby Absur Khan Siam » Sat Jan 28, 2017 11:10 pm

Tasnood's solution(latexed):
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 $\triangle DQC$ and $\triangle FQB$, $\angle Q$ is common and $\angle BFQ = \angle CDQ$ as $AB||CD$ and $QD$ is bisector. So, $\triangle DQC$~$\triangle FQB$.
Then, $FB:CD=QB:QC=1:8$
$\angle QFB = \angle AFD$, $AD||BC||QC$, $\angle ADF = \angle FQB$, $\triangle AFD$~$\triangle BFQ$. $QB:AD=BF:AF=1:7$.
$\frac{QB}{AD} \times \frac{AD}{ED}= \frac{1}{7} \times \frac{4}{3}$
or, $\frac{QB}{ED} = \frac{4}{21}$
Again, $AD:ED=BC:ED=4:3$
So, $\frac{QB}{ED} + \frac{BC}{ED} = \frac{4}{21} + \frac{4}{3}$
or, $\frac{QB+BC}{ED} = \frac{4+4×7}{21}$
or, $\frac{QC}{ED} = \frac{32}{21}$
We know, $\triangle QPC$ ~ $\triangle EPD$ as $AD||BC||QC||ED$.
So, $CP:PE = QC:ED = 32:21$.
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid
Absur Khan Siam
 
Posts: 53
Joined: Tue Dec 08, 2015 4:25 pm
Location: Bashaboo , Dhaka


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