Difference between revisions of "2023 WSMO Speed Round Problems/Problem 1"
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The left triangle has an area of <math>\tfrac{1}{2}\cdot1\cdot2 = 1</math>, the center parallelogram has an area of <math>2\cdot2 = 4</math>, and the right triangle has an area of <math>\tfrac{1}{2}\cdot1\cdot2 = 1</math>. Thus, our answer is <math>1+4+1 = \boxed{6}.</math> | The left triangle has an area of <math>\tfrac{1}{2}\cdot1\cdot2 = 1</math>, the center parallelogram has an area of <math>2\cdot2 = 4</math>, and the right triangle has an area of <math>\tfrac{1}{2}\cdot1\cdot2 = 1</math>. Thus, our answer is <math>1+4+1 = \boxed{6}.</math> | ||
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~pinkpig | ~pinkpig | ||
Latest revision as of 11:05, 12 September 2025
Problem
Find the number of square units in the area of the shaded region.
![[asy] size(4cm); for (int i=0; i<5; ++i) { for (int j=0; j<5; ++j) { dot((i,j)); } } pair a = (0,2); pair b = (1,4); pair c = (3,2); pair d = (4,4); pair e = (3,0); pair f = (1,2); path x = a--b--c--d--e--f--cycle; fill(x, cyan); draw(x, linewidth(2)); [/asy]](http://latex.artofproblemsolving.com/4/2/7/427c11501207e88c8cf5f9f22fd42090a3b5ac30.png)
Solution
We divide the shaded region into three sections as follows
![[asy] size(4cm); for (int i=0; i<5; ++i) { for (int j=0; j<5; ++j) { dot((i,j)); } } pair a = (0,2); pair b = (1,4); pair c = (3,2); pair d = (4,4); pair e = (3,0); pair f = (1,2); path x = a--b--c--d--e--f--cycle; fill(x, cyan); draw(x, linewidth(2)); draw(b--f,red+linewidth(2)); draw(c--e,red+linewidth(2)); [/asy]](http://latex.artofproblemsolving.com/5/9/5/5959a6b89314180367276c12f2582e3a28212297.png)
The left triangle has an area of 
, the center parallelogram has an area of 
, and the right triangle has an area of . Thus, our answer is 
~pinkpig
