National BDMO Secondary P8

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Kazi_Zareer
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National BDMO Secondary P8

Unread post by Kazi_Zareer » Tue Jan 24, 2017 8:09 pm

$\triangle ABC$ is inscribed in circle $\omega$ with $AB = 5$, $BC = 7$, $AC = 3$. The bisector of $\angle A$ meets side $BC$ at $D$ and circle $\omega$ at a second point $E$. Let $\gamma$ be the circle with diameter $DE$. Circles $\omega$ and $\gamma$ meet at $E$ and at a second point $F$. Then $AF^{2} = \frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$.
We cannot solve our problems with the same thinking we used when we create them.

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Kazi_Zareer
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Re: National BDMO Secondary P8

Unread post by Kazi_Zareer » Tue Jan 24, 2017 8:16 pm

I solved by using angle bisector theorem, stewart's theorem, power of the point, law of cosines etc. I think I just messed up but got an answer $AF^2 = 900/19$. So, $m + n = 900 + 19 = 919$

Anyone?
We cannot solve our problems with the same thinking we used when we create them.

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Tasnood
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Re: National BDMO Secondary P8

Unread post by Tasnood » Thu Feb 08, 2018 8:47 pm

Please help to solve:
Secondary-10,2016.PNG
Secondary-10,2016.PNG (46.67 KiB) Viewed 1675 times
In $\triangle ABC$, $AD$ is an angle bisector. So, $\frac{BD}{CD}=\frac{AB}{AC} \Rightarrow \frac{BD}{CD}=\frac{5}{3} \Rightarrow \frac{BC}{CD}=\frac{8}{3} \Rightarrow CD=\frac{3 \times 7}{8} \Rightarrow CD=\frac{21}{8}$
So, $BD=BC-CD=7-\frac{21}{8}=\frac{35}{8}$
Applying Stewart's Theorem, we get:
$BC(AD^2+BD.CD)=AC^2.BD+AB^2+CD \Rightarrow 7AD^2+7\frac{21 \times 35}{8 \times 8}=3^2.\frac{35}{8}+5^2.\frac{21}{8}$
So,$AD=\frac{15}{8}$

In $\triangle ABC$, using cosine law for $\angle BAC$, we get:
$BC^2=AB^2+AC^2-2AB.AC \times cos \angle BAC \Rightarrow 7^2=5^2+3^2-2.5.3 \times cos \angle BAC$
We get: $cos \angle BAC=-\frac{1}{2}$. So, $\angle BAC=120$. So, $\angle BAD= \angle CAD=60$
On arc $CE$, $\angle CAE= \angle CBE=60$. On arc $BE$, $\angle BAE= \angle BCE=60$ So, $\angle BEC=60$ and, $\triangle BEC$ is equilateral.
$BC=BE=CE=7$ Applying Euler's Theorem, we get:
$AB \times CE+AC \times BE=AE \times BC \Rightarrow 5 \times 7+3 \times 7=7AE \Rightarrow AE=8$. So,$DE=AD-AE=8-\frac{15}{8}=\frac{49}{8}$

prottoydas
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Re: National BDMO Secondary P8

Unread post by prottoydas » Tue Mar 13, 2018 9:13 pm

[After Tasnood]
In the cyclic $BCFE$ quadrileteral,we get $/angleBFC=120$.$/angleDFE=90$ so,$/angleCFD=30$.Now $FD$ bisects $/angleBFC$
then
https://artofproblemsolving.com/communi ... 80p2644107
[it is a well known geometry problem from AIME 2012.So,sad that it is th 8th problem in 2012 national problem.

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samiul_samin
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Re: National BDMO Secondary P8

Unread post by samiul_samin » Tue Mar 13, 2018 9:49 pm

prottoydas wrote:
Tue Mar 13, 2018 9:13 pm
[After Tasnood]
In the cyclic $BCFE$ quadrileteral,we get $/angleBFC=120$.$/angleDFE=90$ so,$/angleCFD=30$.Now $FD$ bisects $/angleBFC$
then
https://artofproblemsolving.com/communi ... 80p2644107
[it is a well known geometry problem from AIME 2012.So,sad that it is th 8th problem in 2012 national problem.
Actually 2016 National Olympiad is a showcase of well known problems.Some other problems were also very well known.But,this problem was really different and tough one(for whom who didn't see it before the exam).

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