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Difference between revisions of "2024 SSMO Relay Round 2 Problems/Problem 3"

(Created page with "==Problem== Let <math>T = TNYWR.</math> A point <math>P</math> is randomly chosen inside the square with vertices <math>A = (0,0), B = (0,T), C = (T,T),</math> and <math>D =...")
 
 
Line 4: Line 4:
  
 
==Solution==
 
==Solution==
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Let <math>P = (x,y).</math> We have
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\begin{align*}
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\sqrt{x^2+y^2}+\sqrt{(X-T)^2+(y-T)^2}&\ge\sqrt{x^2+(y-T)^2}+\sqrt{(x-T)^2+y^2}\implies\\
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x^2+y^2+(x-T)^2+(y-T^2)&+2\sqrt{x^2+y^2}\cdot\sqrt{(X-T)^2+(y-T)^2}\ge\\ x^2+(y-T)^2+(x-T)^2&+y^2+2\sqrt{x^2+(y-T)^2}\cdot\sqrt{(x-T)^2+y^2}\implies\\
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(x^2+y^2)((x-T)^2+(y-T)^2)&\ge(x^2+(y-T)^2)(y^2+(x-T)^2)\implies\\
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y^2(x-T)^2+(x^2)(y-T)^2&\ge x^2y^2+(x-T)^2(y-T)^2\implies\\
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(x^2-(x-T)^2)(y^2-(y-T)^2)&\le0\implies\\
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(T^2-2xT)(T^2-2yT)&\le0\implies\\
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\left(x-\frac{T}{2}\right)\left(y-\frac{T}{2}\right)\le0.
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\end{align*}
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So, if we split the square into four smaller squares by drawing lines <math>x=\frac{T}{2},y = \frac{T}{2},</math> the set <math>S</math> is equivalent to the bottom left and top right squares. Thus, the perimeter of the set <math>S</math> is equivalent to <math>4T = 4\cdot890 = \boxed{3560}.</math>
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 +
~SMO_Team

Latest revision as of 14:44, 10 September 2025

Problem

Let $T = TNYWR.$ A point $P$ is randomly chosen inside the square with vertices $A = (0,0), B = (0,T), C = (T,T),$ and $D = (T,0)$. Find the perimeter of the set $S$ containing all points $P$ for which $AP + CP \ge BP + DP.$

Solution

Let $P = (x,y).$ We have \begin{align*} \sqrt{x^2+y^2}+\sqrt{(X-T)^2+(y-T)^2}&\ge\sqrt{x^2+(y-T)^2}+\sqrt{(x-T)^2+y^2}\implies\\ x^2+y^2+(x-T)^2+(y-T^2)&+2\sqrt{x^2+y^2}\cdot\sqrt{(X-T)^2+(y-T)^2}\ge\\ x^2+(y-T)^2+(x-T)^2&+y^2+2\sqrt{x^2+(y-T)^2}\cdot\sqrt{(x-T)^2+y^2}\implies\\ (x^2+y^2)((x-T)^2+(y-T)^2)&\ge(x^2+(y-T)^2)(y^2+(x-T)^2)\implies\\ y^2(x-T)^2+(x^2)(y-T)^2&\ge x^2y^2+(x-T)^2(y-T)^2\implies\\ (x^2-(x-T)^2)(y^2-(y-T)^2)&\le0\implies\\ (T^2-2xT)(T^2-2yT)&\le0\implies\\ \left(x-\frac{T}{2}\right)\left(y-\frac{T}{2}\right)\le0. \end{align*} So, if we split the square into four smaller squares by drawing lines $x=\frac{T}{2},y = \frac{T}{2},$ the set $S$ is equivalent to the bottom left and top right squares. Thus, the perimeter of the set $S$ is equivalent to $4T = 4\cdot890 = \boxed{3560}.$

~SMO_Team

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