Indian Mathematical Olympiad 1988 - Problem 8

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Indian Mathematical Olympiad 1988 - Problem 8

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Leon

#1 h Aug 14, 2009, 12:08 PM • 2 Y

Y by Adventure10, Mango247

A river flows between two houses $A$ and $B$ , the houses standing some distances away from the banks. Where should a bridge be built on the river so that a person going from $A$ to $B$ , using the bridge to cross the river may do so by the shortest path? Assume that the banks of the river are straight and parallel, and the bridge must be perpendicular to the banks.

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Bugi

#2 h Aug 14, 2009, 12:35 PM • 2 Y

Y by Adventure10, Mango247

Translate A into C by vector $\vec{PQ}$ , where PQ is the bridge, with Q and B on one side of the river

Then the path is $AP+PQ+QB=AC+CQ+QB\ge AQ+QB$ , with equality when A,C,Q are collinear.

Construction: Construct C as described, one end of the bridge is intersection of straight path between C and B, other is easily derived.

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#3 h Dec 1, 2010, 4:15 AM • 1 Y

Y by Adventure10

let the river flows along y axis.

see where ever we create the bridge the person must go through a path (a polygonal path) which consists of three line segments of which the middle one has a constant length(wrt to the position of the bridge).
so enough to minimize the total length of the rest two segments.
it is then equivalent to the following question
we can squeeze the river so that two ends of the bridge coincides.then what is the shortest path from A to B ?
it is the line joining A and B.
let the perpendicular distance from river bank to A is $\ h_{A}$
let the perpendicular distance from river bank to A is $\ h_{B}$
let the horizontal distance between A and B is $\ h$
then consider a straight line starting from A which is inclined towards B (wrt x-axis) at an angle $\theta$
where $\theta= arctan(\frac{h}{h_{A}+h_{B}})$
let that line intersect the river bank at the point C.
create the bridge at the point C.

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#4 h Dec 1, 2010, 11:59 AM • 2 Y

Y by Adventure10, Mango247

See the drawing to understand my notations:

$l, a, b, X$ and $Y$ are constants. The distance changes as soon as m and n changes.
Assume that the mininal path is such that this relation is valid: $\frac{m}{a}=\frac{Y}{a+b}=\frac{n}{b}$

To prove it's mininal let's assume the oposite, there's a path shorter than the one I showed:

$\sqrt{a^2+(\frac{aY}{a+b})^2}+l+\sqrt{b^2+(\frac{bY}{a+b})^2}>\sqrt{a^2+m^2}+l+\sqrt{b^2+n^2} \Rightarrow \sqrt{(a+b)^2+Y^2}>\sqrt{a^2+m^2}+\sqrt{b^2+n^2} \\ \Rightarrow a^2+b^2+m^2+n^2+2ab+2mn>a^2+b^2+m^2+n^2+2\sqrt{a^2+m^2}\sqrt{b^2+n^2} \Rightarrow ab+mn> \sqrt{a^2+m^2}\sqrt{b^2+n^2} \\ \Rightarrow a^2 b^2+m^2 n^2+2abmn>a^2 b^2+m^2 n^2 + a^2 n^2+b^2 m^2 \Rightarrow (an-bm)^2<0$

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