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Geometry Difficulty 6.0 National Olympiad Prove it Canada

Problem:
Consider all line segments of length 44 with one end-point on the line y=xy = x and the other end-point on the line y=2xy = 2x. Find the equation of the locus of the midpoints of these line segments.

Solution

Solution:
Let the end-points of the segment be A(a,a)A(a, a) on y=xy = x and B(b,2b)B(b, 2b) on y=2xy = 2x.

The length of ABAB is 44:
AB=(ba)2+(2ba)2=4 AB = \sqrt{(b - a)^2 + (2b - a)^2} = 4
So,
(ba)2+(2ba)2=16 (b - a)^2 + (2b - a)^2 = 16
Expanding:
(ba)2+(2ba)2=(ba)2+(4b24ab+a2)=(ba)2+4b24ab+a2 (b - a)^2 + (2b - a)^2 = (b - a)^2 + (4b^2 - 4ab + a^2) = (b - a)^2 + 4b^2 - 4ab + a^2
But (2ba)2=4b24ab+a2(2b - a)^2 = 4b^2 - 4ab + a^2, so:
(ba)2+(2ba)2=(b22ab+a2)+(4b24ab+a2)=5b26ab+2a2=16 (b - a)^2 + (2b - a)^2 = (b^2 - 2ab + a^2) + (4b^2 - 4ab + a^2) = 5b^2 - 6ab + 2a^2 = 16
Let the midpoint MM of ABAB be (a+b2,a+2b2)\left(\frac{a + b}{2}, \frac{a + 2b}{2}\right).
Let x=a+b2x = \frac{a + b}{2} and y=a+2b2y = \frac{a + 2b}{2}.

Then:
a=2xb a = 2x - b
y=a+2b2=(2xb)+2b2=2x+b2 y = \frac{a + 2b}{2} = \frac{(2x - b) + 2b}{2} = \frac{2x + b}{2}
So,
2y=2x+b    b=2y2x 2y = 2x + b \implies b = 2y - 2x
Substitute bb into aa:
a=2x(2y2x)=2x2y+2x=4x2y a = 2x - (2y - 2x) = 2x - 2y + 2x = 4x - 2y
Now, substitute aa and bb in the length equation:
5b26ab+2a2=16 5b^2 - 6ab + 2a^2 = 16
Plug in a=4x2ya = 4x - 2y, b=2y2xb = 2y - 2x:

First, compute b2b^2:
b2=(2y2x)2=4(yx)2 b^2 = (2y - 2x)^2 = 4(y - x)^2
a2a^2:
a2=(4x2y)2=16x216xy+4y2 a^2 = (4x - 2y)^2 = 16x^2 - 16x y + 4y^2
abab:
ab=(4x2y)(2y2x)=4x(2y2x)2y(2y2x)=8xy8x24y2+4xy=(8xy+4xy)8x24y2=12xy8x24y2 ab = (4x - 2y)(2y - 2x) = 4x(2y - 2x) - 2y(2y - 2x) = 8x y - 8x^2 - 4y^2 + 4x y = (8x y + 4x y) - 8x^2 - 4y^2 = 12x y - 8x^2 - 4y^2
But let's expand directly:
ab=(4x2y)(2y2x)=4x2y4x2x2y2y+2y2x=8xy8x24y2+4xy=(8xy+4xy)8x24y2=12xy8x24y2 ab = (4x - 2y)(2y - 2x) = 4x \cdot 2y - 4x \cdot 2x - 2y \cdot 2y + 2y \cdot 2x = 8x y - 8x^2 - 4y^2 + 4x y = (8x y + 4x y) - 8x^2 - 4y^2 = 12x y - 8x^2 - 4y^2
Now, plug into the equation:
5b26ab+2a2=54(yx)26(12xy8x24y2)+2(16x216xy+4y2) 5b^2 - 6ab + 2a^2 = 5 \cdot 4(y - x)^2 - 6(12x y - 8x^2 - 4y^2) + 2(16x^2 - 16x y + 4y^2)
Compute each term:
5b2=20(yx)25b^2 = 20(y - x)^2
6ab=6(12xy8x24y2)=72xy+48x2+24y2-6ab = -6(12x y - 8x^2 - 4y^2) = -72x y + 48x^2 + 24y^2
2a2=32x232xy+8y22a^2 = 32x^2 - 32x y + 8y^2

Sum:
20(yx)2+48x2+24y272xy+32x232xy+8y2=16 20(y - x)^2 + 48x^2 + 24y^2 - 72x y + 32x^2 - 32x y + 8y^2 = 16
But 20(yx)2=20(y22xy+x2)=20y240xy+20x220(y - x)^2 = 20(y^2 - 2x y + x^2) = 20y^2 - 40x y + 20x^2
So sum all x2x^2 terms:
20x2+48x2+32x2=100x220x^2 + 48x^2 + 32x^2 = 100x^2
y2y^2 terms: 20y2+24y2+8y2=52y220y^2 + 24y^2 + 8y^2 = 52y^2
xy-x y terms: 40xy72xy32xy=144xy-40x y - 72x y - 32x y = -144x y

So the equation is:
100x2+52y2144xy=16 100x^2 + 52y^2 - 144x y = 16
Or, dividing both sides by 44:
25x2+13y236xy=4 25x^2 + 13y^2 - 36x y = 4
Thus, the equation of the locus is:
25x2+13y236xy=4 25x^2 + 13y^2 - 36x y = 4

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