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 =...") |
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==Solution== | ==Solution== | ||
| + | Let <math>P = (x,y).</math> 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 <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> | ||
| + | |||
| + | ~SMO_Team | ||
Latest revision as of 14:44, 10 September 2025
Problem
Let 
A point 
is randomly chosen inside the square with vertices 
and 
. Find the perimeter of the set 
containing all points for which 
Solution
Let 
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 
the set is equivalent to the bottom left and top right squares. Thus, the perimeter of the set is equivalent to 
~SMO_Team
