Minimizing segments of a triangle

A

annajee

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I have already worked this out, but I'm not sure how I did. I think I did this stuff right, but I'd really like to make sure. I really would appreciate someone looking at it and telling me what they think.

Mathematicians from around the world have met and decided to create the Knowbel Prize for Mathematics. The committee charged with giving the award has asked you to help in the creation of the plaque for the winner. Their specifications are that it be in the shape of an isosceles triangle 30 cm on the base with a height of 30 cm. The triangle will be made of silver and placed on a mahogany board with a 2 cm border all around. The crowning achievement will be the valuable diamonds, all of the same shape and size, placed on the three lines AD, BD, and CD inside the triangle, where the point D is equidistant from the points A and B. That is, the lines AD, BD, and CD will all have diamonds mounted on them, without any space between the diamonds. This is where you come into the picture. Your task will be to find the location of point D inside the triangle so that the sum of the lengths of those three segments is minimal. This would insure that the cost of the diamonds is minimal. Since these are mathematicians, they will expect you to write out all of your work in a clear and concise manner, explaining your procedure in detail and showing the mathematics used. In addition they want to know the distances AD, BD, and CD, and the three central angles around D, in degrees.
There is a picture of it, but I don't know how to get that in the email here so I am going to have to try to just describe it in words, so here goes: A triangle (like one side of a pyramid) The very top peak of the triangle is point C, the left bottom corner is point A, the right bottom corner is point B. Inside the triangle is point D. Point D LOOKS like it is about 1/3 of the way up the the height of the triangle. And it is in the center between the two sides of the triangle. So, let me describe it this way: If I put the triangle on a coordinate axes with point A at the origin - then it looks from the picture like point D would be at about (15,10). The line CD goes straight down from the very top (point C) ABOUT 2/3 way down the triangle to point D. The line AD goes from the left bottom corner (point A) at what looks like it could be close to a 30 degree angle, up to point D. The line BD goes from the right bottom corner (point B), at what also looks like about a 30 degree angle, up to point D.



Here's what I've tried and I'm not sure if it's right or not, but I think it is but I was having some trouble working it out completely.
Can you just tell me if I'm on the right track here?

I put the triangle on cartesian coordinates, with the point A at the origin. So then I know that point A is at (0,0), point B is at (30,0), point C is at (x,30). I believe x=15 for point C and point D. From looking at the picture, and because of symmetry. Well, I also know that point D is equidistant from points A and B, so I believe that means that D has to be right in the middle, which would be x=15. So then point C is at (15,30), and point D is at (15,y).
With that info I wrote an expression for the sum of the three lengths (AD, CD, BD) using the distance formula:

f(y)=2*(sqrt(15^2+y^2)) + sqrt((30-y)^2)

I need to minimize so: I took the derivative of that and got:
2y
f'(y)= --------------- -1
sqrt(225 + y^2)

I need to solve for y, so I set f'(y) (which I hope I calculated correctly)
set f'(y)=0

2y
---------------- -1 =0
sqrt(225 + y^2)

For some reason I have been having some trouble solving for y. But I got this answer finally that I will show you how I got it. Hopefully that's right.

2y
--------------- =1
sqrt(225 + y^2)


multiply each side by: sqrt(225 + y^2)


2y = sqrt(225 + y^2)

Divide both sides by 2

sqrt(225 + y^2)
y = ---------------
2

Don't know what to do with that.

How about:

2y = sqrt(225 + y^2)

square both sides

4y^2= 225 + y^2

Subtract y^2 from both sides

4y^2 - y^2 = 225

3y^2 =225

Divide both sides by 3

y^2 = 75

Take the sqrt of both sides

y= + or - (sqrt(75))

has to be positive sqrt(75)

sqrt(75)= 8.660254038

so:

y = 8.660254038

Point D is at (15,8.660254038)

So now I can find the distances for AD, BD, and CD.

distance of AD = sqrt((15-0)^2 + (8.660254038 - 0)^2)
=17.32050808

distance of BD = sqrt((30 - 15)^2 + (0 - 8.660254038)^2)
=17.32050808

distance of CD = 30 - 8.660254038 = 21.33974596

Now to find the three central angles around D in degrees:

Dividing the bottom triangle (AD,BD,AB) in half on x=15,
I can use tan(theta)=8.660254038/15 to find the angle at inside that bottom triangle at point A:
tan(theta)=0.5773502692

So I used arctan to find theta:
arctan(0.5773502692)=30 degrees

With that information, I know that the angle in that bottom triangle at point B is also 30 degrees. Then I just took 180 - 60(the sum of those two angles) =120
So the angle in the triangle below point D at D is 120 degrees.

Then I drew a circle around point D (around those three angles).
Circle=360 degrees.
So: 360-120(that I just found)=240 degrees.
That is the sum of the other two angles around point D.
And those two angles are equal so: 240/2 gives the angle for one: 120 degrees.
So all three angles around point D are 120 degrees.

Does all of this look ok? Do you think these are all right answers and all?
 
Hello, annajee!

I would put it on the graph like this . . .
Code: 
                C
                p(0,30)
               /|\
              / | \
             /  |  \ 
            /   oD  \
           /  * | *  \ 
          / *   |   * \
         /*     |     *\
      A o - - - + - - - o B
    (-15,0)     E    (15,0)

\(\displaystyle \text{Point }D\text{ has coordinates }(0,y).\)

\(\displaystyle \text{The distances are: }\;\begin{Bmatrix}DA &=& \sqrt{u^2+225} \\ DB &=& \sqrt{y^2 + 225} \\ DC &=& 20 - y \end{Bmatrix}\)

\(\displaystyle \text{The total distance is given by: }\;Z \;=\;2(y^2+225)^{\frac{1}{2}} + 30 - y\)

\(\displaystyle \text{Differentiate and equate to zero: }\;Z' \;=\;2\cdot\frac{1}{2}(y^2+225)^{-\frac{1}{2}}(2y) - 1 \;=\;0\)

\(\displaystyle \text{We have: }\;\frac{2y}{\sqrt{y^2+225}} \;=\;1 \quad\Rightarrow\quad 2y \:=\:\sqrt{y^2+225} \quad\Rightarrow\quad 4y^2 \:=\:y^2+225\)

. . \(\displaystyle 3y^2 \:=\:225 \quad\Rightarrow\quad y^2 \:=\:75 \quad\Rightarrow\quad\boxed{ y \:=\:5\sqrt{3}}\)


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

\(\displaystyle \text{In right triangle }DEB\text{, we have: }\:DE = 5\sqrt{3}.\;EB = 15\)
. . \(\displaystyle \text{Then Pythagorus says: }\:DB = 10\sqrt{3}\)

\(\displaystyle \text{We find that }\Delta DEB\text{ is a 30-60 right triangle.}\)
. . \(\displaystyle \text{Hence: }\;\angle DBE = 30^o,\;\angle BDE = 60^o\)

\(\displaystyle \text{Therefore: }\;\angle ADB \:=\:\angle BDC \:=\:\angle CDA \:=\:120^o\)

 
Thank you very much for your input. But I'm wondering: Does it matter if you move the triangle over to -15? Isn't the way I did it acceptable? Cause I got all the same answers.
 
annajee said:
... Isn't the way I did it acceptable? Cause I got all the same answers.
Click to expand...


Hello Anna:

You asked if the way that you did not do the exercise is acceptable.

Since all of your answers match, I would say that the way in which you did do the exercise is acceptable.

Cheers,

~ Howard I. Noe :)

 
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