CPhill's Challenge

Here's an old one that I ran across the other day.....I still think it's a nice problem.....

As seen below, a line is tangent to the parabola y = x^2 at  C

At the same time, a line with the same slope cuts the parabola at AB

Your mission, should you decide to accept it, is to prove that the area ADCBA [ the area between the segment AB and the "bottom"  part of the parabola ] is 4/3 that of the area of triangle ABC.........Good Luck....!!!!

We have:

$\boxed{~ \text{parabola}: \quad y = x^2 \qquad y' = 2x \\ \text{line } \overline{AB} : \quad y = m\cdot x +b\\ \vec{A} = \dbinom{x_a}{y_a} = \dbinom{x_a}{x_a^2}\\ \vec{B} = \dbinom{x_b}{y_b} = \dbinom{x_b}{x_b^2}\\ \vec{C} = \dbinom{x_c}{y_c} = \dbinom{x_c}{x_c^2} ~}$

1. Area of triangle ABC

$\begin{array}{rcl} 2A_{\text{triangle}} = 2A_t &=& | (\vec{C} - \vec{A})\times (\vec{B} - \vec{A})|\\ &=& \left| \dbinom{x_c-x_a}{x_c^2-x_a^2} \times \dbinom{x_b-x_a}{x_b^2-x_a^2} \right| \\ &=&(x_c-x_a)(x_b^2-x_a^2)-(x_c^2-x_a^2)(x_b-x_a)\\ &=&(x_c-x_a)(x_b-x_a)(x_b+x_a)-(x_c-x_a)(x_c+x_a)(x_b-x_a)\\ &=&(x_c-x_a)(x_b-x_a)[(x_b+x_a)-(x_c+x_a)]\\ &=&(x_c-x_a)(x_b-x_a)(x_b+x_a-x_c-x_a)\\ \mathbf{2A_t}& \mathbf{=}&\mathbf{(x_c-x_a)(x_b-x_a)(x_b-x_c)}\\ \end{array}$

$\begin{array}{rcl} \text{slope of the line } \overline{AB}: \quad 2x_c = m &=&\frac{y_b-y_a}{x_b-x_a}\\ &=&\frac{x_b^2-x_a^2}{x_b-x_a}\\ &=&\frac{(x_b-x_a)(x_b+x_a)}{x_b-x_a}\\ &=&x_b+x_a\\ 2x_c = m &=&x_b+x_a\\ 2x_c &=& x_b+x_a\\ \boxed{~x_c = \frac{x_b+x_a}{2} ~} \end{array}\\\\ \begin{array}{rcl} \\ x_c-x_a &=& \frac{x_b+x_a}{2}-x_a = \frac{x_b-x_a}{2}\\ x_b-x_c &=& x_b - \frac{x_b+x_a}{2} = \frac{x_b-x_a}{2}\\\\ 2A_t& =& (x_c-x_a)(x_b-x_a)(x_b-x_c) \qquad |\qquad x_c-x_a = x_b-x_c = \frac{x_b-x_a}{2}\\ 2A_t& =& (\frac{x_b-x_a}{2})(x_b-x_a)(\frac{x_b-x_a}{2} )\\ 2A_t& =& \frac{1}{4}\cdot (x_b-x_a)^3\\ \boxed{~ A_t = \frac{1}{8}\cdot (x_b-x_a)^3 ~} \end{array}$

2. Area of Parabola(Button) ADCBA

$\begin{array}{rcl} A_{\text{parabola}} = A_p &=& \int \limits_{x_a}^{x_b} { (mx+b)\ dx} - \int \limits_{x_a}^{x_b} { x^2\ dx}\\ &=& m\int \limits_{x_a}^{x_b} { x\ dx} + b\int \limits_{x_a}^{x_b} { dx} - \int \limits_{x_a}^{x_b} { x^2\ dx}\\ &=& \frac{m}{2}[x^2]_{x_a}^{x_b} + b[x]_{x_a}^{x_b} - \frac{1}{3}[x^3]_{x_a}^{x_b}\\ \mathbf{A_p} & \mathbf{=} & \mathbf{ \frac{m}{2}(x_b^2-x_a^2) + b(x_b-x_a) - \frac{1}{3}(x_b^3-x_a^3) }\\\\ && \boxed{~ \begin{array}{rcl} (a-b)^3 &=& a^3-3a^2b+3ab^2-b^3\\ a^3-b^3&=&(a-b)^3+3ab(a-b)\\ a^3-b^3&=&(a-b)[(a-b)^2+3ab]\\ a^3-b^3&=&(a-b)[a^2-2ab+b^2+3ab]\\ a^3-b^3&=&(a-b)[a^2+ab+b^2]\\ \end{array} ~}\\\\ && x_b^3-x_a^3 = (x_b-x_a)(x_b^2+x_bx_a+x_a^2)\\\\ A_p & = & \frac{m}{2}(x_b-x_a)(x_b+x_a) + b(x_b-x_a) - \frac{1}{3}(x_b-x_a)(x_b^2+x_bx_a+x_a^2) \\ A_p & = & (x_b-x_a)[\frac{m}{2}(x_b+x_a) + b - \frac{1}{3}(x_b^2+x_bx_a+x_a^2) ] \\\\ && \text{line }: \quad y=mx+b \qquad m=\ ? \qquad b=\ ? \\ && \boxed{~ \begin{array}{rcl} \frac{ y-y_a } { x - x_a } &=& \frac{ y_b-y_a } { x_b - x_a }\\ y-y_a &=& (x - x_a) \frac{ y_b-y_a } { x_b - x_a }\\ y &=& (x - x_a) \frac{ y_b-y_a } { x_b - x_a } +y_a \quad y_b = x_b^2 \quad y_a = x_a^2\\ y &=& (x - x_a) \frac{ x_b^2-x_a^2 } { x_b - x_a } +x_a^2\\ y &=& (x - x_a) \frac{ (x_b-x_a)(x_b+x_a) } { x_b - x_a } +x_a^2\\ y &=& (x - x_a)(x_b+x_a) +x_a^2\\ y &=& x(x_b+x_a) - x_a(x_b+x_a) +x_a^2\\ y &=& x(x_b+x_a) - x_ax_b-x_a^2 +x_a^2\\ y &=& x\underbrace{(x_b+x_a)}_{=m} \ \underbrace{- x_ax_b}_{=b}\\ \end{array} ~}\\\\ \mathbf{m}&\mathbf{=}& \mathbf{x_b+x_a}\\ \mathbf{b}&\mathbf{=}& \mathbf{-x_ax_b}\\\\ A_p & = & (x_b-x_a)[\frac{x_b+x_a}{2}(x_b+x_a) -x_ax_b - \frac{1}{3}(x_b^2+x_bx_a+x_a^2) ] \quad |\quad \cdot \frac66\\ A_p & = & \frac16(x_b-x_a)[3(x_b+x_a)^2 -6x_ax_b - 2(x_b^2+x_bx_a+x_a^2) ] \\ A_p & = & \frac16(x_b-x_a)(3x_b^2+6x_bx_a+3x_a^2 -6x_ax_b - 2x_b^2-2x_bx_a-2x_a^2) \\ A_p & = & \frac16(x_b-x_a)(x_b^2 -2x_bx_a+ x_a^2) \\ A_p & = & \frac16(x_b-x_a)(x_b-x_a)^2 \\ \boxed{~ A_p = \frac16(x_b-x_a)^3 ~}\\ \end{array}$

3. Ratio $\frac{A_t}{A_p}$

$\begin{array}{rcl} \frac{A_t}{A_p} &=& \frac { \frac{1}{8}\cdot (x_b-x_a)^3 } { \frac16(x_b-x_a)^3 }\\ \frac{A_t}{A_p} &=& \frac{1}{8}\cdot \frac61 \\ \frac{A_t}{A_p} &=& \frac{6}{8} \\ \mathbf{\dfrac{A_t}{A_p}} &\mathbf{=} & \mathbf{\dfrac{3}{4} }\\ \end{array}$

Omi67 's answer

http://web2.0calc.com/questions/the-solution-way-with-an-example/new#edit

(not general so Chris did not accept it     )

I liked it!

Great work Heureks    

Very nicely done heureka!

Very nice, heureka...I re-worked this one the other day....I did something similar with the integration part, but, for the area of the triangle, I found the length of the base and then used the "formula" for finding the distance from a point not on a line [in this case, C]  to the line segment AB to find the altitude of the triangle......

There are probably several other methods to solving this.......I actually found it in a different way a long time ago, but, now........I don't remember exactly what I did.....LOL!!!!!

Good job....!!!....5 points from me.....!!!!