Area of this triangle?

Question

Challenge problem:

Consider the polynomial
(x-7)(x-24)(x-25)-19
Construct a triangle whose sidelengths are the roots of this polynomial. What is the area of the resulting triangle? Justify your answer.

Exact value, please. No approximations!

Mathgirl: Those are not the right roots!
But you are going in the right direction.

The triangle is not a right triangle.

Answer 1

Let's say the roots are a,b,c so that the polynomial is:

(x-a)(x-b)(x-c) = x^3 - (a+b+c)x^2 + (ab+bc+ca)x - abc

which must be the same as:

(x-7)(x-24)(x-25) - 19 = x^3 - 56x^2 + 943x - 4219

So we can identify:

C1 = a+b+c = 56
C2 = ab+bc+ca = 943
C3 = abc = 4219

The square of the area of the triangle using Heron's formula can be written in the form:

16 * Area^2 =
(a+b+c)*
[bc^2+ac^2+cb^2+ab^2+ca^2+ba^2
-c^3-b^3-a^3-2abc]

The problem is now to try to express the above expression in terms of C1,C2,C3. After trying to form various expressions made from C1,C2,C3 (since all such functions are symmetric under all permutations of a,b,c, there fortunately aren't that many different types of terms of each degree), I finally came up with:

16 * Area^2
= C1(4C2*C1 - 8C3 - C1^3)
= 4C2*C1^2 - 8C1*C3 - C1^4
= 4*943*56^2 - 8*56*4219 - 56^4
= 104384
==>
A = sqrt(6524)

I see mathgirl has priority with this answer (but I didn't think we were supposed to do numerical root finding?)

Answer 2

I don't want to multiply that out, then try to solve the resulting cubic. But without putting that kind of effort into it, I notice that 7-24-25 is a right triangle. 7, 24, and 25 are roots of the polynomial, leaving out the 19.

Now -- just a thought -- if we could get each of those binomials to equal the cube root of 19, we'd have a solution. Thus

x = 7 + 19^(1/3)
x = 24 + 19^(1/3)
x = 25 + 19^(1/3)

are the roots. The problem is that the resulting triangle has no right angle. There's a formula for the area of a scalene triangle where only the sides are given. The formula (from memory) is something like A = s(s-a)(s-b)(s-c) where a, b, and c are the sides, and s = (a + b + c)/2.

But I'm not sure of that, and don't want to do the work.

I think I could also get the area using trig functions, but again, that's too much effort for now.

What puzzles me is where the 19 came from. The other three numbers form a Pythagorean triple, and somehow I suspect the 19 is related to one also, but I don't see the connection.

Addendum: I cranked numbers, and got roots of 9.668, 26.668, and 27.668. Using the earlier poster's area formula, s = 32.00, and A = 128.517.

I'm not going to check my work.


Addendum 2. I don't know why I punish myself. Mathematician wants an exact solution. So here goes.

Let C = (19^(1/3))/2 (half the cube root of 19).
Then 2C = 19^(1/3).

The roots of the polynomial are

a = 7 + 19^(1/3) = 7 + 2C = 2C + 7
b = 24 + 19^(1/3) = 24 + 2C = 2C + 24
c = 25 + 19^(1/3) = 25 + 2C = 2C + 25

Let s = (a + b + c)/2 = (6C + 56)/2 = 3C + 28

Then s - a = C + 21
s - b = C + 4
s - c = C + 3

In Heron's formula

s(s - c) = (3C + 28)(C + 3) = 3C^2 + 37C + 84 **1**
(s - a)(s - b) = (C + 21)(C + 4) = C^2 + 25C + 84 **2**

When you multiply (1) and (2), you get a quartic

3C^4 + 112C^3 + 1261C^2 +5208C + 7056

if I didn't make a mistake. But C^3 = 19/8, and C^4 = 19C/8,
so 3C^4 + 112C^3 = 57C/8 + 266. After clearing the fraction (multiplying through by 8), the quartic becomes

(10088C^2 + 41721C + 58576)/8 **3**

The area of the triangle is the square root of that thing. It's not a trinomial square, as far as I can tell, and C = (19^(1/3))/2 = 1.3342... is irrational.

When I plug that stuff into my calculator, I get an area of 128.5485, which agrees (within rounding) with my earlier result.

For what it's worth, the solution is expression (3).

But what's the point?

Answer 3

The answer is 2*sqrt(1631).

There has to be an easier way than the way I did it, though! It starts with multiplying the polynomial out, to get

x^3 - 56x^2 + 943x - 4219

and noticing that, if the three roots are r1, r2, and r3, then

r1 + r2 + r3 = 56,
r1r2 + r1r3 + r2r3 = 943, and
r1r2r3 = 4219.

Multiplying those things in various combinations, adding and subtracting the results, and using a form of Hero's Formula:

A = 1/4 sqrt(2((r1r2)^2 + (r1r3)^2 + (r2r3)^2) - (r1^4 + r2^4 + r3^4))

gives A = 1/4 sqrt(104384) = 2*sqrt(1631).

So... very messy, but... there it is! I'd love to see a "prettier" answer.

Answer 4

this triangle is a right triangle.

roots of above given polynomial can be seen as three sides and putting those values in pythagoras formula its

25^2= 24^2+7^2

so height is 24 and base is 7
so the area should be

(1/2)(24)(7)= 84

Answer 5

The roots are
7.0625380413227, 23.310507137328 and 25.626954821349
Using the formula
A = sqrt(s(s-a)(s-b)(s-c)), where a, b and c are the side lengths and s is the semiperimeter,
this gives the area as
A = sqrt(6524)
A = 2 sqrt(1631)
A = 80.771282025188

Answer 6

I havent done this in along time

sorry forgot

Answer 7

Pi


just kiddin'. It's been too long!