Jump to content
BrainDen.com - Brain Teasers
  • 0

minimum distance


BMAD
 Share

Question

13 answers to this question

Recommended Posts

  • 0

For those who can't access this image:

There is a triangle BPC where on one line segment BC there is a point D such that BD=2 and DC=3. Extending through P is a line segment going from some point A to D which is a distance of 5 and perpendicular to BC. Find the minimum total distance of all the "cables" by placing p in an optimal place.

Edited by BMAD
Link to comment
Share on other sites

  • 0

The point is placed ~0.099917m or about 10cm from the bottom. Too lazy to write up the equations, basically used calculus to find the minimum (can never get enough calculus ;P)

The height of A really doesn't matter, since that term disappears when you take the derivative, which makes sense, if you think about it.

Link to comment
Share on other sites

  • 0

Agree with Y-san's answer.

Avoiding tortuous trig derivatives:

Put P at (2, y).

  1. sin B + sin C = 1 (Condition for extremum is dy/dBP + dy/dCP = dy/dy)
  2. tan B / tan C = 3/2 (Inspection)

Guess B reasonably, say 45o.

Solve [1] for C. Check [2].

Increment B by small amount and repeat.

Update B based on change.

Repeat a few times.

B = 35.12633457oC = 25.12653497o

y = 2 tan B = 1.406997295

Link to comment
Share on other sites

  • 0

I got the same figure.

But, with plasmid, I don't know where the cables are.

Assumption (from post 1): We are to minimize PA + PB + P

yes. PA + PB + PC = L

Actually you're right, the answer is 1.407, apparently my algebra with square roots sucks ;P

I am not sure what 1.407 is referring to but i get a different answer.

  • Downvote 1
Link to comment
Share on other sites

  • 0

The height of point P, i.e. the distance between P and D.

The total distance of rope or whatever was used is, as a function of the height of P, which is denoted x:

f(x)=sqrt(x^2+4)+sqrt(x^2+9)+(5-x)

Taking the derivative is pretty simple:

f'(x)=x/sqrt(x^2+4)+x/sqrt(x^2+9)-1

Set f'(x)=0 to find the minimum, and after a giant mess of algebra (which apparently I suck at :/) or plugging it into a numerical solver ;), you'll get a root of x=1.407.

Link to comment
Share on other sites

  • 0

The height of point P, i.e. the distance between P and D.

The total distance of rope or whatever was used is, as a function of the height of P, which is denoted x:

f(x)=sqrt(x^2+4)+sqrt(x^2+9)+(5-x)

Taking the derivative is pretty simple:

f'(x)=x/sqrt(x^2+4)+x/sqrt(x^2+9)-1

Set f'(x)=0 to find the minimum, and after a giant mess of algebra (which apparently I suck at :/) or plugging it into a numerical solver ;), you'll get a root of x=1.407.

I see the issue, I was looking for L the total minimal length of PA,PB,PC while you found the length of DP. Nice job.

Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Answer this question...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Loading...
 Share

  • Recently Browsing   0 members

    • No registered users viewing this page.
×
×
  • Create New...