HMMT, 2012

#Algebra#TrigonometricSubstitution#HMMT#NMTC

Problem Statement

Let $\alpha $ and $ \beta$ be real numbers. The least possible value of

$$ (2 \cos{\alpha}+ 5 \sin{\beta} - 8 )^2 + (2 \sin{\alpha}+ 5 \cos{\beta} -15 )^2 $$

One way to attack this problem could be to expand all the terms and see if any idea from trigonometric optimisation is possible. But that seems unwieldy. So, let us try something else. Say that:

$$ 2 \cos{\alpha}+ 5 \sin{\beta} - 8 = x, \ \text{and} \ 2 \sin{\alpha}+ 5 \cos{\beta} -15 = y $$

We have to minimise $(x^2+y^2)$ under the constraint that:

$$ (x+8) = 2 \cos{\alpha}+ 5 \sin{\beta}, \ \text{and} \ y+15 = 2 \sin{\alpha}+ 5 \cos{\beta} $$

$$ \implies (x+8)^2 + (y+15)^2 = ( 2 \cos{\alpha}+ 5 \sin{\beta})^2 + ( 2 \sin{\alpha}+ 5 \cos{\beta})^2 $$

$$ \implies (x+8)^2 + (y+15)^2 = 29 + 20 \sin{(\alpha+\beta)} \le 49. $$

So, we have to minimise the function $(x^2+y^2)$ under the constraint that: $ (x+8)^2 + (y+15)^2 \le 49. $

So, we have to find the minimum distance of the point from the origin which lies on or inside the circle $ (x+8)^2+ (y+15)^2 = 49. $ The distance of $(0,0)$ from $(-8, -15 )$ is $17$ and the radius of the circle is $7$. Therefore, the required distance is $17- 7 =10.$

AXS

Bibliography & References

[1]
HMMT, 2012, NMTC-1, 2026