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. Given an isosceles trapezoid $ABCD$ with $A(0,0)$, $B(a,0)$, $C(\frac{a+c}{2},h)$ and $D(\frac{a-c}{2},h)$. On the axis of symmetry of this trapezoid: $P(\frac{a}{2},t)$, t is a parameter. The direction of the line $\left[AP\right]$ is: \[m_{AP}=\frac{t-0}{\frac{a}{2}-0}=\frac{2t}{a}\] The direction of the line $\left[PD\right]$ is: \[m_{PD}=\frac{t-h}{\frac{a}{2}-\frac{a-c}{2}}=\frac{2(t-h)}{c}\] Both legs of the trapezoid subtend right angles at $P$ if \[m_{AP}\cdot m_{PD} = -1\] \[\frac{2t}{a} \cdot \frac{2(t-h)}{c}=-1\] \[2t \cdot 2(t-h) = -ac\] \[t^{2}-h\cdot t +\frac{ac}{4} = 0\] The discriminant of this quadratic polynomial: \[\Delta=h^{2}-a \cdot c\] 1. \[\Delta=0\Rightarrow h^{2}=a \cdot c\ \Rightarrow h=\sqrt{a \cdot c}\] Hence $t=\frac{h}{2}$, only one point P, in the middle. 2. \[\Delta > 0 \Rightarrow t_{1}=\frac{h+\sqrt{\Delta}}{2}\ \ \wedge \ \ t_{2}=\frac{h-\sqrt{\Delta}}{2}\] There are two points P.

Attachments:

1.If h>(ac)^.5 we have two points P and if h=(ac)^.5 we have only one point P and for h<(ac)^.5 none points P 2.The distances of P from base AB are: h/2-(h^2-ac)^.5/2 and h/2+(h^2-ac)^.5/2

Surprised that little posts on this IMO problem. Here’s my construction: Let the trapezoid be ABCD with AD=BC. Let E, F, H and I be the midpoints of AB, CD, AD, and BC, respectively, and G be the intersection of HI with EF. Take the circle with diameter AD, and find where it intersects with the axis of symmetry (this is the connection of midpoints of AB and CD since it is symmetrical). Those are the two points P, since they lie on the circle and inscribed angle of a semicircle, satisfying a). As for part b), first by symmetry we note all of the right angles (I trust you can see them ). (a>c) We have $(a-c)^2/4+h^2=AD^2=(2AH)^2=4HJ^2, HG=(a+c)/4$. This means $FJ’^2=EJ^2=(EG-JG)^2=(h/2-JG)^2, JG=\sqrt{HJ^2-HG^2}=\sqrt{(a-c)^2/16+h^2/4-(a+c)^2/16}=\sqrt{\frac{ac+h^2}{4}}; EJ^2=(h/2-\frac{\sqrt{ac+h^2}}{2})^2$. Because of a shortage of time, I won’t simplify this, but hopefully there were no computation mistakes. I had the diagram in GeoGebra but I accidentally closed the tab and now it’s gone. P only exists when the circle intersects with the line of symmetry, which occurs when $\sqrt{(a-c)^2/16+h^2/4}=AH=r=HP\geq (a+c)/4$, or equivalently when $h^2\geq ac$. Whew! Finally, $\blacksquare$