Question

Square $ABCD$ has center $O,$ $AB = 900,$ $E$ and $F$ are on $\overline{AB}$ with $AE \lt BF$ and $E$ between $A$ and $F,$ $m\angle EOF = 45^\circ,$ and $EF = 400.$ Given that $BF = p + q\sqrt{r},$ where $p,$ $q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime, find $p + q + r.$

Answer

Solution:

Let $G$ be the midpoint of $\overline{AB},$ so $OG \perp AB$ and $OG = 450.$ With $\alpha = \angle EOG$ and $\beta = \angle FOG$ on either side of ray $OG,$ we have $EG = 450\tan\alpha,$ $FG = 450\tan\beta,$ and $\alpha + \beta = 45^\circ.$ From $EG + FG = EF = 400,$ we get $\tan\alpha + \tan\beta = \frac{8}{9}.$

The tangent addition formula gives $1 = \tan 45^\circ = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta},$ so $\tan\alpha\tan\beta = 1 - \frac{8}{9} = \frac{1}{9}.$ Hence $\tan\alpha$ and $\tan\beta$ are the roots of $9t^2 - 8t + 1 = 0,$ namely $\frac{4 \pm \sqrt{7}}{9}.$

Since $AE = 450 - EG$ and $BF = 450 - FG,$ the condition $AE \lt BF$ means $EG \gt FG,$ so $\tan\beta = \frac{4 - \sqrt{7}}{9}.$ Then $BF = 450 - 450 \cdot \frac{4 - \sqrt{7}}{9} = 250 + 50\sqrt{7},$ and $p + q + r = 250 + 50 + 7 = 307.$

2005 AIME II Problem 12

Solution:

Problem 12 in Other Years