MHB What is the minimum perimeter of a triangle with given vertices?

[Ackbach]

Here is this week's POTW:

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Given a point $(a,b)$ with $0<b<a$, determine the minimum perimeter of a triangle with one vertex at $(a,b)$, one on the $x$-axis, and one on the line $y=x$. You may assume that a triangle of minimum perimeter exists.

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[Ackbach]

Re: Problem Of The Week # 245 - Dec 19, 2016

This was Problem B-2 in the 1998 William Lowell Putnam Mathematical Competition.

Congratulations to Opalg and IanCg for their correct solutions. Here is Opalg's solution: $+1$ for cool TikZ picture.

Spoiler

[TIKZ]\coordinate (X) at (5,1.5) ;\coordinate [label=left: $A$] (A) at (2,2) ;\coordinate [label=below left: $B$] (B) at (3.25,0) ;\draw (-0.5,0) -- (5.5,0) ;\draw (0,-1.5) -- (0,5.5) ;\draw (0,0) -- (5,5) ;\draw (B) -- (X) -- (1.5,5) -- (A) -- (B) -- (5,-1.5) -- (X) -- (A) ;\draw (6,1.5) node {$P = (a,b)$} ;\draw (6.25,-1.5) node {$Y = (a,-b)$} ;\draw (2,5.25) node {$X = (b,a)$} ;\draw [blue,dashed] (1.5,5) -- (5,-1.5) ;\draw [red] (X) -- (2.725,2.725) -- (4.2,0) -- cycle ;[/TIKZ]​

Let $P = (a,b)$ be the given point. Let $X = (b,a)$ be the reflection of $P$ in the line $y=x$, and let $Y = (a,-b)$ be the reflection of $P$ in the $x$-axis. Let $A$ be a point on $y=x$ and let $B$ be a point on the $x$-axis.

The triangles $PAX$ and $PBY$ are isosceles, so that $PA = AX$ and $PB = BY$. Therefore the perimeter of the triangle $PAB$ is the sum of the lengths $XA$, $AB$, $BY$. This is obviously minimised when $XABY$ is a straight line.

So move $A$ and $B$ to the points on their respective lines where these lines intersect the line $XY$. The corresponding triangle (shown in red in the diagram) will be the one with minimal perimeter. This perimeter is the distance $d$ from $X$ to $Y$, given by $d^2 = (b-a)^2 + (a+b)^2 = 2(a^2 + b^2)$.

Conclusion: the minimal perimeter is $\sqrt{2(a^2+b^2)}$.