## geomerty

For students of class 6-8 (age 12 to 14)
Posts: 29
Joined: Mon Jan 23, 2017 10:32 am

### geomerty

plz, help me solving this
Attachments
16358130_1662673394026350_328482358_o.jpg (79.85 KiB) Viewed 3189 times

Absur Khan Siam
Posts: 65
Joined: Tue Dec 08, 2015 4:25 pm
Location: Bashaboo , Dhaka

### Re: geomerty

Let's extend AC and define two points $F$ and $E$ such that $CE = BC$ and
$CD = CF$.
By some angle chasing,
$\angle DEF = \angle EDF \Rightarrow DF = EF$
$\angle DFA = \angle FAD \Rightarrow DF = AD$
$\therefore AD = EF$

Now ,
$AD + CD = EF + CF = CE = BC = 2017$
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid

jayon_2
Posts: 7
Joined: Sat Feb 11, 2017 4:06 pm

### Re: geomerty

Absur Khan Siam wrote:Let's extend AC and define two points $F$ and $E$ such that $CE = BC$ and
$CD = CF$.
By some angle chasing,
$\angle DEF = \angle EDF \Rightarrow DF = EF$
$\angle DFA = \angle FAD \Rightarrow DF = AD$
$\therefore AD = EF$

Now ,
$AD + CD = EF + CF = CE = BC = 2017$
Summation of any two side of a triangle is bigger than the third side.
But u have showed that,$AD + CD = EF + CF = CE = BC = 2017$,Here $$AD+CD=BC$$ which is practically impossible.?!

aritra barua
Posts: 57
Joined: Sun Dec 11, 2016 2:01 pm

### Re: geomerty

<FAD itself measures 100 degree,then how can you concurr that AD=DF?I think you did not consider triangle DAF. ..According to your construction,<BCE=140 as it is supplementary to <ACB. Therefore,as CF=CD, <FCD=160 and <DFC=<DFA merely measures 10...check out your angle chasing...

Thamim Zahin
Posts: 98
Joined: Wed Aug 03, 2016 5:42 pm

### Re: geomerty

This is a Dhaka regional 2017. I solved it using scale and compass(just measured the length with my scale. Got $2017$, which was definitely the answer because this is 2017) :p
I think we judge talent wrong. What do we see as talent? I think I have made the same mistake myself. We judge talent by the trophies on their showcases, the flamboyance the supremacy. We don't see things like determination, courage, discipline, temperament.

dshasan
Posts: 66
Joined: Fri Aug 14, 2015 6:32 pm

### Re: geomerty

I am showing the angle chasing part

Extend $AC$ in such a way so that $F,E$ are on the opposite side of $C$. Now, $\angle CFD = \angle CDF = 80$
$\angle FDA = 20$, $\angle DAF = \angle DFA = 80$, so, $DA = DF$.
Again, $\angle CBE = \angle CEB = 70$, $\angle DBE = 30$. Now, extend $CD$ so that it meets $BE$ at $X$. $\angle CXE = 90$.Now, $\bigtriangleup CXE$ is congruent to $\bigtriangleup CXB$. So, $BX = XE$. SO, $\bigtriangleup BDE$ is isosceles and $\angle DEB = 30 \Rightarrow \angle DEF = 40$. So, $\bigtriangleup DEF$ is isosceles and $DF = FE$. So, $CD + AD = CF + DF = CF + EF = CE = BC = 2017$.
The study of mathematics, like the Nile, begins in minuteness but ends in magnificence.

- Charles Caleb Colton

Thanic Nur Samin
Posts: 176
Joined: Sun Dec 01, 2013 11:02 am

### Re: geomerty

Here is an easy solution:

Take $K$ on $BC$ so that $CD=CK$. Now, since $\triangle CDK$ is isosceles and $\angle DCK=20^{\circ}$, $\angle DKC=80^{\circ}$ implying $ADKC$ is cyclic.

Now, $\angle BDK=\angle BCA=\angle ABC$, so $BK=DK$. Again, $CD$ is the angle biscector, so $AD=DK$. So $AD+CD=BK+KC=BC=2017$
Hammer with tact.

Because destroying everything mindlessly isn't cool enough.

Atonu Roy Chowdhury
Posts: 63
Joined: Fri Aug 05, 2016 7:57 pm

### Re: geomerty

The simplest solution seems to me:

1. Use Cosine Law to find $AB$
2. Angle Bisector Theorem to find $AD$
3. Cosine Law on $\triangle ADC$ to find $CD$
This was freedom. Losing all hope was freedom.

Posts: 181
Joined: Mon Mar 28, 2016 6:21 pm

### Re: geomerty

What is cosine law?
Frankly, my dear, I don't give a damn.

Atonu Roy Chowdhury
Posts: 63
Joined: Fri Aug 05, 2016 7:57 pm
In $\triangle ABC$ , $c^2 = a^2 + b^2 - 2ab \cos C$