Maths Olympiad Prep

Library / /3 of 11

Geometry Difficulty 4.0 AMC 10/12 Find the answer China

If one side of square ABCDABCD is on the line y=2x17y = 2x - 17, and the other two vertices lie on parabola y=x2y = x^2. Then the minimum area of the square is ________.

A number or a short expression. Spacing and $ signs are ignored.

Solution

Assume that ABAB is on the line y=2x17y = 2x - 17 and the coordinates of the other two vertices on the parabola are C(x1,y1)C(x_1, y_1) and D(x2,y2)D(x_2, y_2). Then CDCD is on a line LL whose equation is y=2x+by = 2x + b. Combining this with the equation of the parabola, we get x2=2x+bx1,x2=1±b+1x^2 = 2x + b \Rightarrow x_1, x_2 = 1 \pm \sqrt{b+1}. Assume that the length of one side of the square is aa. Then
a2=(x1x2)2+(y1y2)2=5(x1x2)2=20(b+1).(1) \begin{aligned} a^2 &= (x_1 - x_2)^2 + (y_1 - y_2)^2 \\ &= 5(x_1 - x_2)^2 = 20(b+1). \qquad (1) \end{aligned}

Pick a point (6,5)(6, -5) on the line y=2x17y = 2x - 17, and the distance from the point to the line y=2x+by = 2x + b is aa.
Soa=17+b5.(2) \text{So} \qquad a = \frac{|17+b|}{\sqrt{5}}. \qquad (2)
From (1) and (2), we get b1=3b_1 = 3, b2=63b_2 = 63, so a2=80a^2 = 80 or a2=1280a^2 = 1280, and amin2=80a_{\min}^2 = 80.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.