Difference between revisions of "2023 WSMO Speed Round Problems/Problem 1"

Revision as of 11:02, 12 September 2025

Problem

Find the number of square units in the area of the shaded region. $[asy] size(4cm); for (int i=0; i<5; ++i) { for (int j=0; j<5; ++j) { dot((i,j)); } } pair a = (0,2); pair b = (1,4); pair c = (3,2); pair d = (4,4); pair e = (3,0); pair f = (1,2); path x = a--b--c--d--e--f--cycle; fill(x, cyan); draw(x, linewidth(2)); [/asy]$

Solution

We divide the shaded region into three sections as follows

$[asy] size(4cm); for (int i=0; i<5; ++i) { for (int j=0; j<5; ++j) { dot((i,j)); } } pair a = (0,2); pair b = (1,4); pair c = (3,2); pair d = (4,4); pair e = (3,0); pair f = (1,2); path x = a--b--c--d--e--f--cycle; fill(x, cyan); draw(x, linewidth(2)); draw(b--f,red+linewidth(2)); draw(c--e,red+linewidth(2)); [/asy]$

The left triangle has an area of $\tfrac{1}{2}\cdot1\cdot2 = 1$ , the center parallelogram has an area of $2\cdot2 = 4$ , and the right triangle has an area of $\tfrac{1}{2}\cdot1\cdot2 = 1$ . Thus, our answer is $1+4+1 = \boxed{6}.$ ~pinkpig

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Difference between revisions of "2023 WSMO Speed Round Problems/Problem 1"

Revision as of 11:02, 12 September 2025

Problem

Solution

@@ Line 51: / Line 51: @@
 draw(c--e,red+linewidth(2));
 </asy>
+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>
+~pinkpig