# 1992 AJHSME Problems/Problem 6

## Problem

Suppose that $[asy] unitsize(18); draw((0,0)--(2,0)--(1,sqrt(3))--cycle); label("a",(1,sqrt(3)-0.2),S); label("b",(sqrt(3)/10,0.1),ENE); label("c",(2-sqrt(3)/10,0.1),WNW); [/asy]$ means $a+b-c$. For example, $[asy] unitsize(18); draw((0,0)--(2,0)--(1,sqrt(3))--cycle); label("5",(1,sqrt(3)-0.2),S); label("4",(sqrt(3)/10,0.1),ENE); label("6",(2-sqrt(3)/10,0.1),WNW); [/asy]$ is $5+4-6 = 3$. Then the sum $[asy] unitsize(18); draw((0,0)--(2,0)--(1,sqrt(3))--cycle); label("1",(1,sqrt(3)-0.2),S); label("3",(sqrt(3)/10,0.1),ENE); label("4",(2-sqrt(3)/10,0.1),WNW); draw((3,0)--(5,0)--(4,sqrt(3))--cycle); label("2",(4,sqrt(3)-0.2),S); label("5",(3+sqrt(3)/10,0.1),ENE); label("6",(5-sqrt(3)/10,0.1),WNW); label("+",(2.5,-0.1),N); [/asy]$ is

$\text{(A)}\ -2 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 2$

## Solution

The first triangle represents $1+3-4$ The 2nd triangle represents $2+5-6$

Solving the first triangle, we get $0$ Solving the 2nd triangle, we get $1$

Since we have to add the 2 triangles the final answer is $1$, which is $\boxed{D}$.