Two rights make a...

Geometry Level 4

The legs of two non-degenerate right triangles $ABC$ and $DEF$ lie on the boundaries of a quadrant of a set of axes. $\triangle ABC$ has vertices $A\left( m,0 \right)$ and $C\left( 0,n \right)$ and $\triangle DEF$ has vertices $D\left( n,0 \right)$ and $F\left(0,m \right)$ .

Let the point where the hypotenuses of $\triangle ABC$ and $\triangle DEF$ intersect be $R$ . What is the possible range of values for $\angle ARF$ ?

0°<x\le 90°

90°\le x<180°

0°\le x\le 90°

90°\le x\le 180°

0°<x<90°

90°<x\le 180°

0°\le x<90°

90°<x<180°

1 solution

Michael Fuller
May 15, 2015

After sketching the two triangles , we find that there are three cases of where the possible angle we are looking for lies. Cases 1 and 2 are shown in the third picture - depending on whether $m$ or $n$ is the largest, the required angle will be on the inside or the outside of the triangles.

Note that in both cases 1 and 2, as either $m$ or $n$ become large while the other stays very small, the angle will get closer to 90°. However it can never reach 90° because that would mean that the triangles would be degenerate such that they lie on the axes of the plane. Therefore we get $\angle ARF>90°$ .

There is another case however. If $m=n$ , then the triangles are identical by SAS congruence and so for case 3, $\angle ARF \le180°$ .

Together, this gives us $\large \boxed { 90°<x \le 180° }$ .

Two rights make a...

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