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### Snug Circle

The length of the radius of circle *b* is six, and its center is at coordinates (0,6). Circle *a* is tangent both to circle *b* and to the line *y*=0; the two points of tangency are different. The radius of circle *a* is two.

*What are the absolute values of the coordinates of the center of circle a?*

### Congruence Time

The Bicoastal Cafe displays two clocks, one on Los Angeles time and the other on New York time (three hours ahead). At any given time, the hour and minute hands on each clock form some angle, which should always be considered a non-negative number of degrees less than or equal to 180°.

*What is the earliest time in the day in Los Angeles (reckoning the beginning of a day at the stroke of midnight there) that the angles formed by each of the clocks are equal?*

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## Solutions to week 67

**Card T**. Let *E* be the center of circle *a*. A careful diagram like the one at the right seems to indicate that the line *DA* is tangent to circle *a*, but how can we be sure? Well, it would suffice to show that radius *EA* is perpendicular to line *DA*. To this end, let angle *BEA* have measure *x* degrees. Then, since triangle *BEA* is isosceles, angle *BAE* has measure 90 – *x*/2. Also, triangle *BEC* is congruent to *BEA*, so the measure of angle *AEC* is 2*x*. Since triangle *CEA* is isosceles, this means that angle *CAE* has measure 90 – *x*. Taking the difference of the two known angles at *A*, we see that angle *BAC* has measure *x*/2. But triangle *DBA* is congruent to triangle *CBA*, so the measure of angle *BAD* is also *x*/2. Adding up the three known angles at *A*, we see that the measure of angle *EAD* is 90°. Hence, line *DA* interesects circle *a* just **one** time, at *A*.

**Card 61**. Since *BF* is parallel to *CE*, you can create a smaller rectangle by “erasing” quadrilateral *BCEF* and squashing the two remaining triangles together vertically. Since quadrilateral *BCEF* has area 1/7 that of rectangle *ACDF*, the resulting rectangle must have area 6/7 of rectangle *ACDF*. But it has the same length *AF*, so it must have height *AB* **6/7** of the height *AC* of *ACDF*.

## Recent Weeks

**Week 67**: Card 61 & Card T, solutions to Long Haul & Postal Pathways

**Week 66**: Long Haul & Postal Pathways, solutions to Card 129 & Card 94

**Week 65**: Card 129 & Card 94, solutions to Skimpy Schedule & Passing Pennies

**Week 64**: Skimpy Schedule & Passing Pennies, solutions to Seating Arrangement & Stuck on Fairness

**Week 63**: Seating Arrangement & Stuck on Fairness, solutions to Grid Adjustment & Way Outside the Box

Links to all of the puzzles and solutions are on the Complete Varsity Math page.

**Come back next week** for answers and more puzzles.

[asciimathsf]