## IGO 2016 Medium/1

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

### IGO 2016 Medium/1

1. In trapezoid $ABCD$ with $AB \parallel CD$, $w_1$ and $w_2$ are two circles with diameters $AD$ and $BC$, respectively. Let $X$ and $Y$ be two arbitrary points on $w_1$ and $w_2$, respectively. Show that the length of segment $XY$ is not more than half of the perimeter of $ABCD$.

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### Re: IGO 2016 Medium/1

$Claim 1:$ In a trapezoid $ABCD (AB \parallel CD), AB + CD = 2EF $ where $E,F$ are the midpoints of $AD, BC$

$Proof:$ Extend $AD, BC$ so that they meet at $X$. Now, let $XA = a, XB = b, AE = ED = c, BF = FC = d$.

Now, $\frac{AB}{EF} = \frac{a}{a+c}$ and $\frac{EF}{CD} = \frac{a+c}{a+2c}$. From this two equation, we get, \[ AB + CD = 2EF\]

Now, segment $XY$ is largest when $XY$ passes through the midpoints of $AD$ and $BC$. Let $m,n$ be the midpoints of $AD$ and $BC$ respectively. SO, $Xm = \frac{1}{2} AD$ and $Yn = \frac{1}{2} BC$.

Therefore, $XY = Xm + mn + Yn = \frac{1}{2} ( AB + BC + CD + DA)$. For all other $XY$, it is less than half of the perimeter of $ABCD$

$Proof:$ Extend $AD, BC$ so that they meet at $X$. Now, let $XA = a, XB = b, AE = ED = c, BF = FC = d$.

Now, $\frac{AB}{EF} = \frac{a}{a+c}$ and $\frac{EF}{CD} = \frac{a+c}{a+2c}$. From this two equation, we get, \[ AB + CD = 2EF\]

Now, segment $XY$ is largest when $XY$ passes through the midpoints of $AD$ and $BC$. Let $m,n$ be the midpoints of $AD$ and $BC$ respectively. SO, $Xm = \frac{1}{2} AD$ and $Yn = \frac{1}{2} BC$.

Therefore, $XY = Xm + mn + Yn = \frac{1}{2} ( AB + BC + CD + DA)$. For all other $XY$, it is less than half of the perimeter of $ABCD$

The study of mathematics, like the Nile, begins in minuteness but ends in magnificence.

- Charles Caleb Colton

- Charles Caleb Colton