## BDMO 2017 National round Secondary 5

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### BDMO 2017 National round Secondary 5

The circular arcs $AC$ and $BC$ have centers at $B$ and $A$ respectively. There exists a circle tangent to both arcs $AC$ and $BC$ and to the line segment $AB$. The length of the arc $BC$ is $12$. What is the circumference of the circle?
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Thanic Nur Samin
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### Re: BDMO 2017 National round Secondary 5

The length of arc $BC$ is $12$, so the circumference of the circle is $72$.

Now, we apply cartesian coordinate. Let $AB=2k$. Take $A$ as the origin and $AB$ as the $x$-axis. Now, define points.

1. $T$: touchpoint of the small circle and the circle with center $A$.
2. $M$: midpoint of $AB$. It has coordinate $(k,0)$
3. $K$: center of the small circle. Let its coordinate be $(k,t)$
4, $L$: the point $(0,-2k)$. It lies on the circle with center $A$. From homothety, $L,M,T$ are collinear.

Now, the circle with center $A$ has equation $x^2+y^2=4k^2$ and the line $LM$ has equation $2x-y-2k=0$. Solving, we get that $T$ is the point $(8k/5,6k/5)$. However, from homothety, $A,K,T$ are collinear. The equation of $AT$ is $y=3x/4$, so $K$ is $(k,3k/4)$. So the radius of the small circle is $3k/4$, while the radius of the big circles are $2k$. So if we multiply the circumference of the big circles by $\dfrac{3k/4}{2k}=3/8$, we get the circumference of the small circle, which is $72\times \dfrac{3}{8}=27$.
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nahin munkar
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### Re: BDMO 2017 National round Secondary 5

A synthetic solution :

We denote the inscribed circle as $\omega$ , & $P,Q$ be the tangent point of arc $BC$ & $AB$ resp with $\omega$.
We get, $AB=AC=r$ (radii of same circle)
& $AB=BC$ similarly.
$\Longrightarrow AB=AC=BC \Longrightarrow \triangle ABC$ is equilateral $\Longrightarrow \angle B =60^{\circ}$.

So, $\dfrac{60^{\circ}}{360^{\circ}} \times 2\pi r =12 \Longrightarrow r= \dfrac{36}{\pi}$.

$\bullet$ Let, $P'$ be the second point of intersection of $AP$ & $\omega$ .It's easy to see that, $AP$ goes through the center of $\omega$ & $AQ=BQ= \dfrac{r}{2}$.

$\star$ By the extension of POP, we get,
$AQ^2= AP \times AP' \Rightarrow \dfrac{r^2}{4} = r \times AP' \Longrightarrow AP' = \dfrac{9}{\pi} \Longrightarrow PP'= \dfrac{36}{\pi} - \dfrac{9}{\pi} = \dfrac{27}{\pi}$
AS, $PP'$ is the diameter, so radius of $\bigodot \omega$ , namely $r_\omega = \dfrac{27}{2 \pi}.$

SO, the circumference of $\bigodot \omega = 2\pi r_\omega = 2\pi \times \dfrac{27}{2\pi}= 27.$
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### Re: BDMO 2017 National round Secondary 5

nahin munkar wrote:
Wed Feb 15, 2017 12:27 am
A synthetic solution :

We denote the inscribed circle as $\omega$ , & $P,Q$ be the tangent point of arc $BC$ & $AB$ resp with $\omega$.
We get, $AB=AC=r$ (radii of same circle)
& $AB=BC$ similarly.
$\Longrightarrow AB=AC=BC \Longrightarrow \triangle ABC$ is equilateral $\Longrightarrow \angle B =60^{\circ}$.

So, $\dfrac{60^{\circ}}{360^{\circ}} \times 2\pi r =12 \Longrightarrow r= \dfrac{36}{\pi}$.

$\bullet$ Let, $P'$ be the second point of intersection of $AP$ & $\omega$ .It's easy to see that, $AP$ goes through the center of $\omega$ & $AQ=BQ= \dfrac{r}{2}$.

$\star$ By the extension of POP, we get,
$AQ^2= AP \times AP' \Rightarrow \dfrac{r^2}{4} = r \times AP' \Longrightarrow AP' = \dfrac{9}{\pi} \Longrightarrow PP'= \dfrac{36}{\pi} - \dfrac{9}{\pi} = \dfrac{27}{\pi}$
AS, $PP'$ is the diameter, so radius of $\bigodot \omega$ , namely $r_\omega = \dfrac{27}{2 \pi}.$

SO, the circumference of $\bigodot \omega = 2\pi r_\omega = 2\pi \times \dfrac{27}{2\pi}= 27.$
how aq=bq=r/2?

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### Re: BDMO 2017 National round Secondary 5

Ooh I got a cool solution to this. So, $OM+OB = AB$ and $MB=\dfrac{AB}{2}$.

Also by pythagorean theorem, $MB^2+OM^2=OB^2$.

Substituting, $\dfrac{AB^2}{4} + (AB-OB)^2 = OB^2$
Or, $\dfrac{AB^2}{4} + AB^2 - 2\times AB \times OB + OB^2 = OB^2$
Or, $\dfrac{AB^2}{4} + AB^2 - 2\times AB \times OB = 0$
Or, $\dfrac{AB^2}{4} + AB^2 = 2\times AB \times OB$
Or, $\dfrac{AB}{4} + AB = 2 \times OB$
Or, $OB = \dfrac{5AB}{8}$

So, $OM= AB - OB = AB - \dfrac{5AB}{8} = \dfrac{3AB}{8}$
Substitute the value of $AB$ and you get the answer.
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nahin munkar
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### Re: BDMO 2017 National round Secondary 5

Mon Feb 05, 2018 8:53 pm

how aq=bq=r/2?
No problem at all. First read carefully.
nahin munkar wrote:
Wed Feb 15, 2017 12:27 am
A synthetic solution :

We denote the inscribed circle as $\omega$ , & $P,Q$ be the tangent point of arc $BC$ & (line) $AB$ resp with $\omega$.
SO, $AB=r$ &
$AQ=BQ= \dfrac{r}{2}$
(as 'Q' is the midpoint of AB)
# Mathematicians stand on each other's shoulders. ~ Carl Friedrich Gauss

Durjoy Sarkar
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### Re: BDMO 2017 National round Secondary 5

Sun Feb 18, 2018 7:58 am
Ooh I got a cool solution to this. So, $OM+OB = AB$ and $MB=\dfrac{AB}{2}$.

Also by pythagorean theorem, $MB^2+OM^2=OB^2$.

Substituting, $\dfrac{AB^2}{4} + (AB-OB)^2 = OB^2$
Or, $\dfrac{AB^2}{4} + AB^2 - 2\times AB \times OB + OB^2 = OB^2$
Or, $\dfrac{AB^2}{4} + AB^2 - 2\times AB \times OB = 0$
Or, $\dfrac{AB^2}{4} + AB^2 = 2\times AB \times OB$
Or, $\dfrac{AB}{4} + AB = 2 \times OB$
Or, $OB = \dfrac{5AB}{8}$

So, $OM= AB - OB = AB - \dfrac{5AB}{8} = \dfrac{3AB}{8}$
Substitute the value of $AB$ and you get the answer.
elementary one.. I like it

Tasnood
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### Re: BDMO 2017 National round Secondary 5

How to say that, $OM+OB=AB$?

Durjoy Sarkar
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### Re: BDMO 2017 National round Secondary 5

radius 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 $OB$.
$MO=OX$
$BX= BO+OX=BO+MO$

Tasnood
Posts: 73
Joined: Tue Jan 06, 2015 1:46 pm

### 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?