1960 IMO Problems/Problem 7

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An isosceles trapezoid with bases $a$ and $c$ and altitude $h$ is given.

a) On the axis of symmetry of this trapezoid, find all points $P$ such that both legs of the trapezoid subtend right angles at $P$;

b) Calculate the distance of $P$ from either base;

c) Determine under what conditions such points $P$ actually exist. Discuss various cases that might arise.


(a) The intersection of the circle with diameter one of the legs with the axis of symmetry.

(b) Let $x$ be the distance from $P$ to one of the bases; then $h - x$ must be the distance from $P$ to the other base. Similar triangles give $\frac{x}{\frac{a}{2}} = \frac{\frac{c}{2}}{h - x}$, so $x^2 - hx + \frac{ac}{4} = 0$ and so $x = \frac{h \pm \sqrt{h^2 - ac}}{2}.$

(c) When $h^2 \ge ac$.


In our above picture, ABCD is our trapezoid with AB=a and CD=c and its height is 'h'. AF and BE are perpendicular to CD such that AF= BE= h. XY is our axis of symmetry and it intersects with CD at a point O. Point O is our origin of reference whose coordinates are (0,0).

Let our point P be on the axis of symmetry at z distance from the origin O.

The coordinates of the points A,B,C,D,E,F and P are given in the figure.


Slope of the line $PC= (z-0)/(0-c/2) =  -2z/c$ Slope of the line $PB= (z-h)/(0-a/2) = -2(z-h)/a$

Since the leg BC subtends a right angle at P, the angle BPC should be a right angle. This means that the product of the slope of PC and PB is -1.



or $z^2 - zh + ac/4= 0$

Now, solving for z, we get, $z=  [(h + ( h^2 - ac ) ^1/2 ]/2$ and $[(h - ( h^2 - ac ) ^1/2 ]/2$

So, z is the distance of the points from the base CD..

Also the points are possible only when , $h^2 - ac >= 0$.. and doesn't exist for $h^2 -ac <0$

See Also

1960 IMO (Problems)
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