Question number 30. A right triangle has legs of lengths a and b. So here I have drawn an example right triangle with two legs labeled a and b. We have two points, one called P lying on the leg of length a and the second one Q lying on the leg of length b. We also have two points on the hypotenuse labeled K and H and these points lie at the intersection of the hypotenuse with lines perpendicular to the hypotenuse and passing through the points P and Q respectively. So together these three segments have the combined length given here in the problem and we are trying to minimize this sum. So what is the smallest value here of the lengths of these red lines in this setup? So what we can notice right away are some bounds on the length of P and Q by moving the points P and Q to various positions. For example if I let P be here at the right angle and Q at the other two vertices we note that a is the lower bound for PQ and b is the upper bound where we assume of course that a is less than b. So the longer leg is the upper bound and the lower bound is the shorter leg of this triangle. But what we can do to absolutely minimize P and Q is we can let P and Q have length equal to 0 and that would happen if we set P to be identically the point Q. So P and Q can be the same point nothing is preventing us from doing that and then what would happen is if we draw the line perpendicular to the hypotenuse passing through that point here maybe like that we have a 90 degree angle over here and that point of intersection is the point K and also the point H. So this is K which is identically equal to H. So then we know that the segment PK is equal in length to the segment QH and so the quantity we're looking for KP plus PQ plus QH it's just equal to twice KP twice the length of KP and KP is the height taken at the hypotenuse. So let's compute let's now use the area formula so we have one half base times height so one half a times b if b is the base and a is the height and that's equal to one half here we have the hypotenuse so the length let me just call it base times height for now and that is one half the base is the hypotenuse so we use the Pythagorean formula here to compute the hypotenuse and the height is exactly the length of KP. So then we can solve for that by simply dividing we have the length of the segment KP equal to a b divided by the square root of the sum of the squares of a and b and that's exactly half our answer the answer here is given by 2 KP so 2 KP is twice what we had and that is answer C.

right triangle

minimize length

hypotenuse intersection

Pythagorean theorem

area relations