Square root transformation

My friend shared this problem with me and I thought it was very interesting.

A transformation of a first quadrant maps the points $(x, y)$ to $( \sqrt{x}, \sqrt{y})$ . The vertices of a quadrilateral ABCD are $A(900, 300), B(1800, 600), C(600, 1800), D(300, 900).$ Find the area of its image under the said transformation.

#Geometry #MathProblem #Math

Easy Math Editor

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`2 \times 3`	$2 \times 3$
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`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

Line $AB$ is $y = \dfrac{1}{3}x$ , which maps to $y^2 = \dfrac{1}{3}x^2$ , i.e. $y = \dfrac{1}{\sqrt{3}}x$ .

Line $BC$ is $x+y = 2400$ , which maps to $x^2+y^2 = 2400$ .

Line $CD$ is $y = 3x$ , which maps to $y^2 = 3x^2$ , i.e. $y =\sqrt{3}x$ .

Line $DA$ is $x+y = 1200$ , which maps to $x^2+y^2 = 1200$ .

Convert all of these to polar to get: $\theta = 30^{\circ}$ , $r = 20\sqrt{3}$ , $\theta = 60^{\circ}$ , $r = 20\sqrt{6}$ .

Hence, the new region is exactly a $30^{\circ}$ sector of a doughnut with inner radius $20\sqrt{3}$ and outer radius $20\sqrt{6}$ .

Therefore, the area is $\dfrac{30^{\circ}}{360^{\circ}} \cdot \pi \cdot \left((20\sqrt{6})^2 - (20\sqrt{3})^2 \right) = \boxed{100\pi}$

Jimmy Kariznov - 7 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$