2007 AMC 10B Problem 15

The chain $\angle A = 2\angle B = 3\angle C = 4\angle D$ says that twice $B$, three times $C$ and four times $D$ all equal $A$. So each angle is a fraction of $A$.

1. Let $\angle A = x$. Then
$$\angle B = \frac{x}{2}, \qquad \angle C = \frac{x}{3}, \qquad \angle D = \frac{x}{4}.$$
2. The four angles of a quadrilateral add to $360^\circ$:
$$x + \frac{x}{2} + \frac{x}{3} + \frac{x}{4} = 360.$$
3. Use twelfths to add the fractions: $\frac{12}{12} + \frac{6}{12} + \frac{4}{12} + \frac{3}{12} = \frac{25}{12}$. So
$$\frac{25}{12}\,x = 360.$$
4. Solve:
$$x = 360 \times \frac{12}{25} = \frac{4320}{25} = 172.8.$$
5. Rounded to the nearest whole number, that is $173$.

The answer is $\boxed{\textbf{(D)}\ 173}$.

When several expressions are all set equal, give their common value one name and write everything in terms of it. Then one sum equation finishes the problem. The phrase "rounded to the nearest whole number" warns you that the exact answer is not whole, so do not be surprised by $172.8$.

Avoid fractions by calling the common value $12k$. Then the angles are $A = 12k$, $B = 6k$, $C = 4k$, $D = 3k$. They add to $25k = 360$, so $k = 14.4$ and $A = 12 \times 14.4 = 172.8$.

Let $\angle A = x$. The chain of equalities says $2\angle B = x$, $3\angle C = x$, $4\angle D = x$, so
$$\angle B = \frac{x}{2}, \qquad \angle C = \frac{x}{3}, \qquad \angle D = \frac{x}{4}.$$

The angles of a quadrilateral add to $360^\circ$:
$$x\left(1 + \frac12 + \frac13 + \frac14\right) = x \cdot \frac{12 + 6 + 4 + 3}{12} = \frac{25x}{12} = 360.$$
Hence
$$x = \frac{360 \cdot 12}{25} = \frac{4320}{25} = 172.8,$$
which rounds to $173$.

The answer is $\boxed{\textbf{(D)}\ 173}$.