I assumed the coordinates of P = (h,2) to get the value of DP+PM= $\sqrt { (h-2)^2 +4}+\sqrt{h^2+1}$. Then I differentiated the equation wrt to h to get: $h(\sqrt{h^2+1}) -2\sqrt{h^2+1}+ h\sqrt{h^2+8-4h}$. On equating this expression with 0, I will probably get the answer.

Is there any other~shorter method of solving this?

up vote 5 down vote accepted

The easiest way is to use a reflection. Let $N$ be the symmetric of $M$ with respect to $B$:

enter image description here

Then $DP+PM=DP+PN\geq DN$, and the minumum for $DP+PM$ is achieved when $P=BC\cap DN$.

  • 1
    Let me add: ABN and BPN triangles in this case are similar and since BN=0.5AB, AN=AB+BN=AB+0.5AB=1.5AB, thus the ratio is 1:3 so 3PB=BC. – chx Feb 7 '15 at 21:14

Here is an easy solution that does not use calculus.

Consider the point $E(4,2)$, which is the mirror image of point $D$ in the line $\overline{BC}$. Any path from $M$ to $P$ to $D$ has the same length as a path from $M$ to $P$ to $E$. But obviously the shortest path from $M$ to $E$ is a straight line segment.

So choose $P$ to be on the line segment $\overline{ME}$. This makes $P(2,2/3)$, and the smallest value of $DP+PM$ is $\sqrt{3^2+2^2}=\sqrt{13}$.

(I was just about to add a graphic to this answer, but I see that Jack D'Aurizio has given a slightly different but equivalent answer with a graphic, so I'll refrain.)

The set of all points $P$ in the plane with the same $DP+PM$ is an ellipse with $D$ and $M$ as its foci. The higher its eccentricity (i.e. the flatter and less circle-like the ellipse is), the smaller that sum is. But if you make it too small, the ellipse will no longer intersect the line $BC$. So the minimal situation is the one where the ellipse merely touches $BC$, i.e. has $BC$ as a tangent.

Any light ray emitted from one focus of the ellipse will be reflected in the ellipse in such a way that it ends up in the other focus. Reflection in the ellipse is equivalent to reflection in the tangent at the point where the ray hits the ellipse. So the line $DP$ passes through $M'$, the reflection of $M$ in $BC$. Likewise the ray $MP$ passes through $D'$, the reflection of $D$ in $BC$. So by constructing $M'$ or $D'$ you can easily find $P$.

Figure

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.