Problem 19 · 2012 Math Kangaroo
Stretch
Geometry & Measurement
spatial-reasoning
The clock shown has a rectangular clock face, the hands however move as usual in a constant circular pattern. How big is the distance x of the digits 1 and 2 (in cm), if the distance between the numbers 8 and 10 is given as 12 cm?
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Answer: C — \(4\sqrt{3}\)
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Hint 1 of 2
The numbers sit where rays \(30^\circ\) apart from the centre meet the rectangle; 8 and 10 are the two left corners, so their \(12\) cm gap is the rectangle's height.
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Hint 2 of 2
Put the centre at the origin and find where the rays toward 1 and 2 hit the frame.
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Approach: intersect the 30°-spaced clock rays with the rectangle
- Numbers 8 and 10 are the bottom-left and top-left corners, so the left edge (the height) is \(12\) cm; place the centre at the origin with half-height \(6\).
- The ray to 2 (\(60^\circ\) from straight up) hits the top-right corner \((W, 6)\): from direction \(\left(\tfrac{\sqrt3}{2}, \tfrac12\right)\), reaching \(y=6\) needs \(t=12\), so \(W = 6\sqrt3\) and corner 2 is at \((6\sqrt3, 6)\).
- The ray to 1 (\(30^\circ\) from up) hits the top edge at \(x = 2\sqrt3\), so number 1 is at \((2\sqrt3, 6)\).
- The gap is \(6\sqrt3 - 2\sqrt3 = 4\sqrt3\) cm, choice C.
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