2007 AMC 10B Problem 15

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 $\boxed{\textbf{(D)}\ 173}$.

A chain like $A = 2B = 3C = 4D$ means all four expressions equal one common value; pick that value as the variable and everything else is a fraction of it. Then a single angle-sum equation finishes the job. The phrase "rounded to the nearest whole number" warns that the exact answer is not an integer, so a whole-number result should raise suspicion.

Let the common value be $12k$ to clear denominators: the angles are $12k, 6k, 4k, 3k$, summing to $25k = 360$, so $k = 14.4$ and $\angle A = 12k = 172.8$.

• Misreading the chain as A = 2B = 3C = 4D meaning B = 2A, C = 3A, D = 4A, which gives the wrong (and tiny) angle A = 36.
• Using $180^\circ$ (a triangle's angle sum) instead of $360^\circ$, which gives $86.4$.