[superprismatic]

What is the length of a side of an equilateral

triangle which has a point inside of it that is

at distances 3, 4, and 5 from its vertices?

[Guest]

I'm getting a number around...

6.75

[Guest]

<p>

I called the point inside the triangle (a,b). Letting the length of the equilateral triangle be n, the vertices of the triangle would be (0,0), (0,n), and (n/2, sqrt(3),n/2). WLOG, the simultaneous equations you would get are:</p>

<p>a^2 + b^2 == 25<br />

b^2 + (a - n)^2 == 16<br />

(a - n/2)^2 + (b - n*(sqrt(3)/2))^2 == 9</p>

<p>Solving this (I used an equation solver), gives you a value of n = sqrt(25 + 12sqrt(3)), which is about 6.76643.

</p>

[Guest]

Solved it geometrically using a 2D CAD program. Mark a point A roughly in the center of your proposed drawing then, with center A, draw 3 circles of radius 3, 4 and 5 units. One corner of the triangle must fall on the circumference of each circle.

An equilateral triangle has corner angles of 60°, so draw a line "z" horizontally (angle 0°) and another line "y" at 60° and a third line "x" at angle 120°.

Line x and z meet at point Y on the circumference of the "3" circle. Now, using the CAD program, move both lines x & z together around the circumference, then move line y such that y crosses lines x & z on the circumference of the 4 circle and 5 circle respectively. About three tries are needed.

Now use the 'dimension' command to measure the side of the equilateral triangle formed by x, y & z and it comes out to 6,753.

[plasmid]

Here's an approach that can (in principle) work without resorting to a computer, but in practice the equations get Messy beyond my threshold of tolerability (which was rather high as shown by my Puzzle Land part V problem)

I started the same way as bobpi: Call the length of the sides of the equilateral triangle L, and place the triangle so one vertex is at the origin, one vertex is at (L, 0), and one vertex is at (L/2, L sqrt(3/4)).

If a line of length 5 is drawn from the vertex at the origin, and a line of length 4 is drawn from the vertex at (L, 0), then we can solve for the y-coordinate where they will intersect (Y): just make use of the Pythagorean theorem and find the height of their intersection point on the y-axis such that the total distance the lines span along the x-axis is equal to L.

sqrt(5^2 - Y^2) + sqrt(4^2 - Y^2) = L

(5^2 - Y^2) + sqrt[(5^2 - Y^2) (4^2 - Y^2)] + (4^2 - Y^2) = L^2

sqrt[(5^2 - Y^2) (4^2 - Y^2)] = L^2 + 2Y^2 - 41

(5^2 - Y^2) (4^2 - Y^2) = L^4 + 4L^2Y^2 - 82L^2 + 4Y^4 - 164Y^2 + 1681

400 - 41Y^2 + Y^4 = 4Y^4 + (4L^2 - 164) Y^2 + (L^4 - 82L^2 + 1681)

3Y^4 + (4L^2 - 123) Y^2 + (L^4 - 82L^2 + 1281) = 0

Y^2 = the quadratic solution of that formula above

This can be solved for Y as a function of L, but the solution is messy. Once you've solved for Y, you can get the x-coordinate of their intersection by just considering the line of length 5.

X = sqrt(5^2 - Y^2)

The the distance D between that point (X, Y) and the last vertex (L/2, L sqrt(3/4)) is

D = sqrt[(X - L/2)^2 + (Y - L sqrt(3/4))^2]

Since Y can be solved as a function of L, and X is a function of Y and therefore also a function of L, then D can be expressed as a function of L. It's then a matter of taking the horrendous equation D = f(L) and manipulating it to solve for L as a function of D, L = g(D), and solving for the case of D = 3.

But hopefully there's a cleaner way of doing this.

[bushindo]

What is the length of a side of an equilateral

triangle which has a point inside of it that is

at distances 3, 4, and 5 from its vertices?

This is such an elegant problem. I'm amazed at how the algebra works out nicely. Thanks

By combining 3 equations for the circles of radii 3, 4, and 5 centered at each of the 3 verteces and Heron's formula, I get the following equation for L

L⁴ - 50 L² + 193 = 0

Solving for L using the quadratic formula, I get

L =[ 25 + .5*( 1728)^1/2 ]^1/2