Difference between revisions of "1960 IMO Problems/Problem 7"
(→Solution) |
|||
Line 11: | Line 11: | ||
(a) The intersection of the circle with diameter one of the legs with the axis of symmetry. | (a) The intersection of the circle with diameter one of the legs with the axis of symmetry. | ||
− | (b)Let <math>x</math> be the distance from <math>P</math> to one of the bases; then <math>h - x</math> must be the distance from <math>P</math> to the other base. Similar triangles give <math>\frac{x}{\frac{a}{2}} = \frac{\frac{c}{2}}{h - x}, so < | + | (b)Let <math>x</math> be the distance from <math>P</math> to one of the bases; then <math>h - x</math> must be the distance from <math>P</math> to the other base. Similar triangles give <math>\frac{x}{\frac{a}{2}} = \frac{\frac{c}{2}}{h - x}</math>, so <math>x^2 - hx + \frac{ac}{4} = 0</math> and so <math>x = \frac{h \pm \sqrt{h^2 - ac}}{2}.</math> |
− | (c) When < | + | (c) When <math>h^2 \ge ac</math>. |
==See Also== | ==See Also== |
Revision as of 00:23, 17 May 2015
Problem
An isosceles trapezoid with bases and
and altitude
is given.
a) On the axis of symmetry of this trapezoid, find all points such that both legs of the trapezoid subtend right angles at
;
b) Calculate the distance of from either base;
c) Determine under what conditions such points actually exist. Discuss various cases that might arise.
Solution
(a) The intersection of the circle with diameter one of the legs with the axis of symmetry.
(b)Let be the distance from
to one of the bases; then
must be the distance from
to the other base. Similar triangles give
, so
and so
(c) When .
See Also
1960 IMO (Problems) | ||
Preceded by Problem 6 |
1 • 2 • 3 • 4 • 5 • 6 • 7 | Followed by Last Question |
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.
Now,
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.
i.e
4z(z-h)=-ac
or z^2 - zh + ac/4= O
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