Triangle Polynomial

Let the side lengths and hence the roots of the polynomial be $a$ , $b$ and $c$ with $c$ being the hypotenuse and hence by Pythagorean theorem $a^2+b^2 = c^2$ . From Vieta's formula , we have $\begin{cases} a+b+c = 20 & ...(1) \\ abc = 17 &...(2) \end{cases}$

From equation (1):

$\begin{aligned} a+b+c & = 20 \\ a+b & = 20 - c & \small \color{#3D99F6} \text{Squaring both sides} \\ {\color{#3D99F6}a^2} + 2ab + {\color{#3D99F6}b^2} & = 400-40c+{\color{#3D99F6}c^2} & \small \color{#3D99F6} \text{Note that }a^2+b^2=c^2 \\ ab & = 200-20c & \small \color{#3D99F6} \text{Multiplying both sides by }c \\ \color{#3D99F6}abc & = 200c -20c^2 & \small \color{#3D99F6} \text{Note that (2): }abc = 17 \\ 20c^2 - 200c + 17 & = 0 & \small \color{#3D99F6} \text{Solving the quadratic for }c \\ \implies c & = \frac {50+\sqrt{2415}}{10} & \small \color{#3D99F6} \frac {50-\sqrt{2415}}{10} \approx 0.0857 \text{ too small for }c \end{aligned}$

The area of the triangle is

$A = \dfrac {ab}2 = \dfrac {ab\color{#3D99F6}c}{2\color{#3D99F6}c} = \dfrac {17}{2 \cdot \frac {50+\sqrt{2415}}{10}} = \dfrac {85(50-\sqrt{2415})}{(50+\sqrt{2415})(50-\sqrt{2415})} = 50-\sqrt{2415}$

$\implies p+q = 50+2415 = \boxed{2465}$

Let $a, b, c$ be the roots of the polynomial. WLOG, suppose $a \leq b < c$ . Since $a, b, c$ are the side lengths of a right triangle, we must have $a^2 + b^2 = c^2$ by the Pythagorean Theorem. Furthermore, from Vieta's, we have

$\begin{aligned} a + b + c &= 20 \\ abc &= 17. \end{aligned}$

The first equation can be written as $a + b = 20 - c,$ while the second equation can be written as $ab = \dfrac{17}{c}.$

Now, we use the fact that $(a + b)^2 = a^2 + b^2 + 2ab.$ Substituting, we can get a quadratic equation in terms of only $c$ :

$\begin{aligned} (a + b)^2 &= a^2 + b^2 + 2ab \\ (20 - c)^2 &= c^2 + 2 \left ( \dfrac{17}{c} \right ) \\ 400 - 40c + c^2 &= c^2 + \dfrac{34}{c} \\ 400 - 40c &= \dfrac{34}{c} \\ 400c - 40c^2 &= 34 \\ -40c^2 + 400c - 34 &= 0 \\ 20c^2 - 200c + 17 &= 0. \end{aligned}$

Applying the quadratic formula (and skipping over all the arithmetic), we get $c = \dfrac{50 \pm \sqrt{2415}}{10}.$

But, there are two possible values for $c$ ! Which one do we choose? (Of course, only one of the choices for $c$ gives the area in the form we want, but this solution makes it more rigorous.)

EDIT: As Wei Chen points out, we can use the fact that $a^2 + b^2 \geq 2ab$ to get $c^2 \geq \dfrac{34}{c}$ , or $c \geq 34^{1/3}$ . This immediately gets rid of the smaller value of $c$ , so we can skip all the proceeding work.

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We can get a hint from our original equations involving $a$ and $b.$ If we substitute our value of $c$ back into these equations, we find that

$\begin{aligned} ab &= \dfrac{17}{\frac{50 \pm \sqrt{2415}}{10}} \\ &= \dfrac{17 \cdot 10}{50 \pm \sqrt{2415}} \\ &= \dfrac{2 \cdot 85(50 \mp \sqrt{2415})}{2500 - 2415} \\ &= 2(50 \mp \sqrt{2415}) \\ \end{aligned}$

and

$\begin{aligned} a + b &= 20 - \dfrac{50 \pm \sqrt{2415}}{10} \\ &= \dfrac{150 \mp \sqrt{2415}}{10} \\ &= \dfrac{100 + 50 \mp \sqrt{2415}}{10}. \end{aligned}$

Let $C = 50 \pm \sqrt{2415},$ where we choose one of either the plus or the minus sign. Now, we have the system of equations

$\begin{aligned} a + b &= \dfrac{C + 100}{10} \\ ab &= 2C. \end{aligned}$

The second equation gives $b = \dfrac{2C}{a}.$ Substituting this into the first equation gives

$\begin{aligned} a + \dfrac{2C}{a} &= \dfrac{C + 100}{a} \\ 10a^2 + 20C &= (C + 100)a \\ 10a^2 - (C + 100)a + 20C &= 0. \end{aligned}$

The solutions to this equation, through the quadratic formula and a little simplification, are $a = \dfrac{(C + 100) \pm \sqrt{C^2 - 600C + 100^2}}{20}.$

If $C = 50 + \sqrt{2415}$ (corresponding to $c = \dfrac{50 - \sqrt{2415}}{10}$ ), then we can say $C \approx 100.$ This shows that $C^2 - 600C + 100^2 < 0,$ so $a$ cannot be a real number. This cannot happen, as $a$ is a side length in a triangle.

However, if $C = 50 - \sqrt{2415}$ (corresponding to $c = \dfrac{50 + \sqrt{2415}}{10}$ ), then $C \approx 0,$ and $C^2 - 600C + 100^2 > 0.$ What's more, because $C$ is slightly above $0,$ $C^2 - 600C + 100^2$ is slightly less than $100^2$ , so $a$ is less than $\dfrac{C + 100 \pm 100}{20},$ which is a real number no matter whether we choose the positive or the negative sign. Similarly, $b$ must also be a real number. Thus, in order to have a right triangle, we must choose $C = 50 - \sqrt{2415}.$

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Going all the way back, we now see that $ab = 2(50 - \sqrt{2415}).$ The area of the right triangle is, therefore, equal to $\dfrac{ab}{2} = 50 - \sqrt{2415}$ , and $p + q = 50 + 2415 = \boxed{2465}.$

Addendum: The right triangle has side lengths

$a = \dfrac{(150 + \sqrt{2415}) + \sqrt{(50 - \sqrt{2415})^2 - 600(50 - \sqrt{2415}) + 100^2}}{20} \approx 9.913 \\ b = \dfrac{(150 + \sqrt{2415}) - \sqrt{(50 - \sqrt{2415})^2 - 600(50 - \sqrt{2415}) + 100^2}}{20} \approx 0.173 \\ c = \dfrac{50 + \sqrt{2415}}{10} \approx 9.914.$

One can check that $a^2 + b^2 = c^2,$ along with all the other equations involving Vieta's.