Perimeter of Triangle in a Right triangle.

I am having difficulty solving this problem:

The perimeter of a right triangle is 18 inches. If the midpoints of three sides are joined by line segments they form another triangle . What is the perimeter of this new triangle ? (Ans: 9 inches) .

Any suggestions on how to solve it ?

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Each side of the new triangle is half the side of the old triangle, meaning that the perimeter is halved as well.

For a little more detail, draw it out. cutting at the midpoint makes this new triangle one-half of a rectangle. So clearly, since you cut at the midpoint, the legs of the new triangle are half the legs of the old, since opposite sides are of equal length in a rectangle. By similar triangles the hypotenuse is halved as well. So the whole perimeter is cut in half and so the answer is $9$.

EDIT: Picture! The rectangle I'm talking about is $AGFE$.

$AE$ is half of the side, so $GF$ is also half of that side. The same argument can be made for $FE$ and $GE$ easily. This argument doesn't need a right angle though: what matters is that you form a parallelogram (opposite sides are of the same length), which always happens. This is just a special case.

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"cutting at the midpoint makes this new triangle one-half of a rectangle". Sounds interesting but i cant quiet picture it. Could you elaborate it please? – MistyD Jun 19 '12 at 17:37
certainly! I've added a picture to demonstrate. – Robert Mastragostino Jun 19 '12 at 17:55
Simple and straightforward.. Thank you – MistyD Jun 19 '12 at 18:01

An expansion of @Robert's way of drawing this out. First, a note: your question is a special case of a Medial Triangle where the outer triangle is specified to be right. That wiki link (and the Wolfram links in the bibliography) give a more complete explanation of the general case.

Anyway, draw your right triangle in such a way that the side lengths sum to 18. It doesn't matter which sides have what lengths so long as they sum to the specified 18 (If you're actually sketching this out on paper, don't even worry about the lengths. This is just to convince you that we're talking about your problem).

Let the vertices be $A$, $B$ and $C$ with $AB$ the hypotenuse. Mark the midpoints of each, with $A'$ the midpoint of the hypotenuse, and $B',C'$ the midpoints of the bases. Note then that $A'C'$ is parallel to $B'C$ and thus $A'B'$ is half the length of $AC$. By a similar argument note that $A'C'$ is half the length of $BC$. Finally, note that $B'C'$ (the hypotenuse of the inscribed triangle) is the hypotenuse of $\triangle C'B'C$ which is similar to $\triangle ABC$ and has bases half the length. Thus, $B'C'$ must be half of $AB$.

Hopefully this helps. It probably works best if you draw it out on graph paper, and are as precise as possible. However, be sure to convince yourself with the argument, not careful drawing and measuring. It may be helpful to think about it as proving the side lengths of the medial triangle are half of the outer triangle, rather than the worry about the specific lengths at first.

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Another nice way to see it is that the old triangle is split into 4 congruent triangles, each of which is similar to the old triangle. Looking at any of the new triangles that shares a vertex with the old, it is clear that two of the sides are half the length of the corresponding sides of the old triangle, so by similarity, the remaining side is again half the length of its corresponding side. By congruence, the same holds for the triangle under discussion, so the perimeter is halved, as desired.

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Hint: the new triangle is similar to the original, as the angles match.

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Note: the fact that the triangle is right doesn't really matter. It just makes it easier to see in a picture. – tomasz Jun 19 '12 at 17:35