Points $A, B, C, D, E$ lie on a circle and a point $P$ lie outside the circle. The given points such that (1) lines $PB$ and $PD$ are tangent to the circle, (2) $P, A, C$ are collinear and (3) $DE$ is parallel to $AC$.
Prove that: $BE$ bisects $AC$
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- Tue Apr 17, 2018 10:17 am
- Forum: Secondary Level
- Topic: Easy Projective Geo
- Replies: 3
- Views: 6726
- Thu Apr 05, 2018 4:56 pm
- Forum: Secondary Level
- Topic: "Log"ing Problem!
- Replies: 1
- Views: 4106
Re: "Log"ing Problem!
$(24log_a2)=a^{128}+128$ $\Rightarrow log_a(log_a2^{24})=a^{128}+128$ $\Rightarrow log_a2^{24}=a^{a^{128}+128}$ [According to the basis of logarithm] $\Rightarrow 2^{24}=a^{a^{a^{128}+128}}$ $\Rightarrow a^{a^{128}.a^{a^{128}}}=2^{24}$ $\Rightarrow y^y=2^{24}$ [Setting $a^{a^{128}}=y$] $\Rightarrow...
- Thu Apr 05, 2018 4:50 pm
- Forum: Secondary Level
- Topic: "Log"ing Problem!
- Replies: 1
- Views: 4106
"Log"ing Problem!
Let $a>1$ and $x>1$ satisfy $log_a(log_a(log_a2)+log_a24-128)=128$ and $log_a(log_ax)=256$.
Find the remainder when $x$ is divided by $1000$
Find the remainder when $x$ is divided by $1000$
- Mon Mar 12, 2018 10:30 am
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Secondary/Higher Secondary 2018/2
- Replies: 1
- Views: 2323
Re: BdMO National Higher Secondary 2018/2
Capture.PNG Let $AC$ and $BD$ meets at $E$. $AD=a,BC=b$ where, $a<b$. Easy to prove: $\angle CAD=\angle ABE, \angle DBC=\angle BAE$ We know, $\angle BAD=\angle BAE+ \angle CAD=90^{\circ}$ $\Rightarrow \angle BAE+\angle ABE=90^{\circ}$ Again, $\angle ABE+\angle ADB=90^{\circ}$ So, $\angle BAE=\angle...
- Sat Mar 10, 2018 12:13 am
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Secondary/Higher Secondary 2018/4
- Replies: 2
- Views: 2796
Re: BdMO National Higher Secondary 2018/4
$21$ I got.
- Wed Mar 07, 2018 10:17 am
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO 2017 National Round Secondary 7
- Replies: 17
- Views: 19788
Re: BdMO 2017 National Round Secondary 7
Told you before, if $20$ pictures can share more than two common color and not all of them share the same color, the answer is $2$.
[100% sure]
[100% sure]
- Wed Mar 07, 2018 10:07 am
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO 2017 National round Secondary 5
- Replies: 15
- Views: 11298
Re: BDMO 2017 National round Secondary 5
You proved $BX=OB+OM$, not $AB=OB+OM$Durjoy Sarkar wrote: ↑Mon Mar 05, 2018 10:06 pmradius of two big circle are same. let X be the point where little circle is tangent.
it is well known the center of little circle, tangent point are lies on $OB$.
$MO=OX$
$BX= BO+OX=BO+MO $
- Mon Mar 05, 2018 11:43 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO 2017 National round Secondary 5
- Replies: 15
- Views: 11298
Re: BDMO 2017 National round Secondary 5
Again a question. How can we say that $A$-centered circle will go through $B$?
- Mon Mar 05, 2018 10:34 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO 2017 National round Secondary 5
- Replies: 15
- Views: 11298
Re: BDMO 2017 National round Secondary 5
Sorry for Interrupt. My answer was same of #Nahin but this solution is new to me.
How can we find the radius of two circles same?
How can we find the radius of two circles same?
- Mon Mar 05, 2018 8:50 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO 2017 National round Secondary 5
- Replies: 15
- Views: 11298
Re: BDMO 2017 National round Secondary 5
How to say that, $OM+OB=AB$?