Triangles between parallel lines

Fix a base BC and a line l parallel to it. Put the third corner A anywhere on l. You get a whole family of triangles.

Their areas are all equal (same base, same height), so there is no biggest and no smallest. That is worth saying clearly, because the next question has a very different answer.

Their perimeters are not equal. Slide A far off to one side and the two slanting sides get huge. Somewhere there is a triangle with the smallest perimeter. Where?

BC is in every triangle of the family, so it is enough to make AB + AC as small as possible.

Here is the trick. Treat the line l as a mirror and reflect C across it, giving C′. From your work on plane mirrors, the image is as far behind the mirror as the object is in front.

So for any A on l, AC = AC′ (the triangles on either side of the mirror are congruent). That means the path B → A → C has exactly the same length as the path B → A → C′, wherever A is on l.

Now the problem is easy. Making B → A → C shortest is the same as making B → A → C′ shortest, and the shortest route from B to C′ is the straight line. So put A where the segment BC′ crosses l. That triangle has the smallest perimeter.

Gopal's problem is the same one in real life. He must walk from his house to the river, fill his pot, then carry it to the water tank. Reflect the tank across the river line, join the house to that image with a straight line, and where it meets the river is the best place to fetch water.