Let $A = (0, 1)$ and $B = (1, 1)$ in the plane $\mathbb{R}^2$ . Determine the length of the shortest path from $A$ to $B$ consisting of the line segments $AP$, $PQ$ and $QB$, where $P$ varies on the $x$-axis between the points $(0, 0)$ and $(1, 0)$ and $Q$ varies on the line $y = 3$ between the points $(0, 3)$ and $(1, 3)$.

I was trying this question many times, but I could not able to solve it. I was using the distance formula, but I didn't get correct value..

If anybody help me I would be very thankful to him..


Let $M(0,-1)$ and $N(1,5)$.

Hence, the length of $MN$ is an answer.

Let $O(0,0)$, $C(0,3)$, $D(1,3)$, $MN\cap CD=\{Q'\}$ and $MN\cap x-axis=\{P'\}.$

Thus, since $DB=DN$ and $AO=NO$, we obtain for all $P$ on $x$- axis and for all $Q\in CD$:

$AP=NP$, $NQ=BQ$, $AP'=NP'$ and $NQ'=BQ'$.

Thus, $$AP+PQ+QB=MP+PQ+QN\geq MP'+P'Q'+Q'N=MN=\sqrt{1^2+6^2}=\sqrt{37}.$$ The equality occurs for $P\equiv P'$ and $Q\equiv Q'$, which says that $\sqrt{37}$ is the answer.

This problem we can prove also by the Minkowski's inequality.

Indeed, let $P(a,0)$ and $Q(b,3)$.

Thus, by Minkowski we obtain: $$AP+PQ+QB=\sqrt{a^2+1}+\sqrt{(a-b)^2+9}+\sqrt{(b-1)^2+4}=$$ $$=\sqrt{a^2+1}+\sqrt{(b-a)^2+9}+\sqrt{(1-b)^2+4}\geq$$ $$\geq\sqrt{(a+b-a+1-b)^2+(1+3+2)^2}=\sqrt{37}.$$


  • $\begingroup$ thanks @Michael Rozenberg $\endgroup$ – lomber Aug 21 '17 at 9:22
  • $\begingroup$ @lomber lego You are welcome! $\endgroup$ – Michael Rozenberg Aug 21 '17 at 9:25

$P\left(\dfrac{1}{6};\;0\right)$ and $Q\left(\dfrac{2}{3};\;3\right)$

The graph is self explanatory. The shortest path is the straight line. I just reflected the given segments in a way such that the path gave a straight line and then reflected again to get back the points on the given segments. Points $M$ and $Q$ are on the same vertical line.

Shortest distance is $r+m+n=\sqrt{1+\dfrac{1}{36}}+\sqrt{\dfrac{1}{4}+9}+\sqrt{4+\dfrac{1}{9}}=\sqrt{37}$

enter image description here

  • $\begingroup$ thanks a lot @Raffaele,,,,,,this is real;;y good explainatory $\endgroup$ – lomber Aug 21 '17 at 9:23

The logic is:- "the shortest distance between two points is the straight line segment between them".

1) (Translate A to its equivalent position.) Find A' = (0, -1), the reflected image of A about the x-axis.

2) (Translate B to its equivalent position.) Find B' = (1, 5), the reflected image of B about the line y = 3.

3) Join A'B', the required line segment.

4) The place that A'B' cuts the x-axis is P. Q is similarly found.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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