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Difference between revisions of "1960 IMO Problems/Problem 7"

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(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.
  
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(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>
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(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 <math>h^2 \ge ac</math>.
 
(c) When <math>h^2 \ge ac</math>.
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==See Also==
 
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{{IMO7 box|year=1960|num-b=6|after=Last Question}}
 
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[[Category:Olympiad Geometry Problems]]
 
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<center>
 
<center>
 
[https://www.artofproblemsolving.com/Wiki/images/0/01/1960_7.jpg]
 
[https://www.artofproblemsolving.com/Wiki/images/0/01/1960_7.jpg]
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Now,  
 
Now,  
  
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Slope of the line PC= (z-0)/(0-c/2) =  -2z/c
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Slope of the line <math>PC= (z-0)/(0-c/2) =  -2z/c</math>
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Slope of the line PB= (z-h)/(0-a/2) = -2(z-h)/a
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Slope of the line <math>PB= (z-h)/(0-a/2) = -2(z-h)/a</math>
  
 
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.  
 
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.  
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i.e  
 
i.e  
  
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4z(z-h)=-ac
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<math>4z(z-h)=-ac</math>
  
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or z^2 - zh + ac/4= O
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or <math>z^2 - zh + ac/4= 0</math>
  
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Now, solving for z, we get,  z=  [(h + ( h^2 - ac ) ^1/2 ]/2  and  [(h - ( h^2 - ac ) ^1/2 ]/2
+
Now, solving for z, we get,  <math>z=  [(h + ( h^2 - ac ) ^1/2 ]/2</math> and  <math>[(h - ( h^2 - ac ) ^1/2 ]/2</math>
  
 
So, z is the distance of the points from the base  CD..  
 
So, z is the distance of the points from the base  CD..  
  
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Also the points are possible only when , h^2 - ac >= 0.. and doesn't exist for h^2 -ac <0
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Also the points are possible only when , <math>h^2 - ac >= 0</math>.. and doesn't exist for <math>h^2 -ac <0</math>
 +
 
 +
==See Also==
 +
 
 +
{{IMO7 box|year=1960|num-b=6|after=Last Question}}
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 +
[[Category:Olympiad Geometry Problems]]

Latest revision as of 14:25, 23 March 2020

Problem

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.

Solution

(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 $h^2 \ge ac$.

[1]

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

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)
Preceded by
Problem 6 		1 • 2 • 3 • 4 • 5 • 6 • 7 Followed by
Last Question
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