 ## Right Triangle Inscribed in a Rectangle with One Leg of Fixed Length Given a rectangle ABCD of fixed dimensions a by b, and a line segment PQ of a fixed length, inscribe a triangle in the rectangle such that one vertex of the triangle is at Point A and P and Q are located on the sides opposite to A such that the triangle APQ is a RIGHT triangle. First issue:

Show that to have triangle APQ to be a right triangle, triangles ABP and PCQ must be similar.

What problems do you encounter when you try for a straightedge and compass (or GSP) construction?

Show that the construction can not be accomplished with straightedge and compass. An approach:

Introduce an x-y coordinate system with the rectangle located  so that point A is at the origin. Label the figure as at the right.

Let x = BP   and  y = DQ.     Our goal is to develop some equations expressing relationships between  x and y.     If we can find two such independent equations then the common solution will allow location of points P and Q . All four triangles are right triangles and the given conditions of a, b, and s will allow use of the Pythagorean theorem to express the lengths of the internal lines. This equation is the equation of a parabola  that passes through points (0,b) and (a,b) and has axis of symmetry at x = ½ b.

Now, the given condition for the length of PQ  is the hypotenuse of  Triangle PCQ.   Therefore we have:  This is the equation for a circle of radius s with a center at (a,b).    So considering together we have an image like that at the right. The result when a = 12, b = 10, and s =5 is shown.   The intersection of  the two curves is at    (8.118, 6.849) and AP =  12.881 approximately.

The fact that the center of the circle is at (a,b) means one of the intersections will be at a point (m,n) such that m > a  and  n > b.   Thus it is outside of the rectangle.

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