MHB How Do You Calculate the Minimal Path Sum in Coordinate Geometry?

  • Thread starter Thread starter Albert1
  • Start date Start date
AI Thread Summary
The minimal path sum in coordinate geometry involves calculating the shortest distance between points P=(3,7) and Q=(1,5) via points A and B. By reflecting point Q across the y-axis and then the x-axis, the shortest distance is determined to be the straight line to the reflected point Q''. The length of the path, denoted as k, equals 4√10. The coordinates for points A and B are found to be A=(2/3,0) and B=(0,2). The final result for x+y+k is 4√10 + 8/3.
Albert1
Messages
1,221
Reaction score
0
P=(3,7),Q=(1,5) ,A=(x,0),B=(0,y)

if k=minimum(PA+AB+BQ)

find: x+y+k
 
Mathematics news on Phys.org
[sp]

Let $Q'$ be the reflection of $Q$ in the $y$-axis and let $Q''$ be the reflection of $Q'$ in the $x$-axis. The shortest distance between $P$ and $Q''$ is the straight line between them. The shortest path between $P$ and $Q$ (via $A$ and $B$) is obtained by reflecting parts of that line in the axes, as appropriate (the red path in the diagram). The length $k$ of the path is the same as the distance from $P$ to $Q''$, which is $4\sqrt{10}.$ The point $A$ is at $(2/3,0)$ and $B$ is at $(0,2)$. So $x+y+k = 4\sqrt{10} + \frac83$.[/sp]
 

Attachments

  • reflections.png
    reflections.png
    4.4 KB · Views: 71
Opalg said:
[sp]
https://www.physicsforums.com/attachments/2267​

Let $Q'$ be the reflection of $Q$ in the $y$-axis and let $Q''$ be the reflection of $Q'$ in the $x$-axis. The shortest distance between $P$ and $Q''$ is the straight line between them. The shortest path between $P$ and $Q$ (via $A$ and $B$) is obtained by reflecting parts of that line in the axes, as appropriate (the red path in the diagram). The length $k$ of the path is the same as the distance from $P$ to $Q''$, which is $4\sqrt{10}.$ The point $A$ is at $(2/3,0)$ and $B$ is at $(0,2)$. So $x+y+k = 4\sqrt{10} + \frac83$.[/sp]
your answer is correct:)
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top