Mathcounts Chapter Competition Solutions Worksheet - Middle School - 2014 Page 5
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11
12DE has radius, BF = √3. Segments BD and
labeled 2 is 2/45, and so on. Thus, the
probabilities of landing on each of the spaces
BE also are radii of the circle containing are
DE. Note that ∆DBE ~ ∆ABC, which means
labeled 3, 5, 7 and 9 are 3/45, 5/45, 7/45
that ∆DBE is an equilateral triangle. It follows,
and 9/45, respectively. It follows that, in a
then, that BD = DE = √3 Ans.
given spin, the probability that the spinner will
land on a space labeled 1, 3, 5, 7 or 9 is
1/45 + 3/45 + 5/45 + 7/45 + 9/45 = 25/45
24. On a number line, how many units apart are
= 5/9 Ans.
the two points that are twice as far from 0 as
they are from 9?
22. The length of the diagonal of rectangle ABCD
Let a be one of the points, and 0 < a < 9. The
is half the rectangle’s perimeter less four-fifths
distances from 0 to a and from a to 9 are a
the length of the shorter side. Find the ratio of
and 9 − a, respectively. Since a is to be twice
the length of the shortest side to the length of
as far from 0 as it is from 9, we can write the
the longest side of the rectangle.
equation 2 × (9 − a) = a. Solving for a, we
Let L and W represent the length and width of
see the first point is 18 − 2a = a → 3a = 18
the rectangle, respectively. Four-fifths of the
→ a = 6. Now, assume the second point, b, is
length of the shortest side of the rectangle is
located on the number line such that b > 9.
(4/5)W.
The distances from 0 to b and from 9 to b are
The rectangle has diagonal of length
b and b − 9, respectively. We have the
√(L
2
+ W
2
) = L + W − (4/5)W
equation 2 × (b − 9) = b. Solving for b, we
√(L
2
+ W
2
) = L + (1/5)W
see the other point is 2b − 18 = b → b = 18.
L
2
+ W
2
= L
2
+ (2/5)LW + (1/25)W
2
It follows that the distance between a and b is
0 = (2/5)LW − (24/25)W
2
18 – 6 = 12 Ans.
(2/5)L = (24/25)W
10L = 24W
3
2
n
25. Given 3n
+ 3n
+ 4n = n
, find n.
5L = 12W
We can simplify this equation a bit by dividing
W/L = 5/12 Ans.
both sides by n:
→ 3n
3
2
n
2
n−1
3n
+ 3n
+ 4n = n
+ 3n + 4 = n
n
n
23. An equilateral triangle has sides of length 2.
That simplifies the equation, but it doesn’t tell
A circular arc is centered at B. Find the
us the value of n. Let’s evaluate the simplified
length of a segment drawn parallel to side AC
equation for several values of n. The results
with endpoints D and E.
Triangle ABC is an equilateral triangle, so
are displayed in the table below.
angle B measures 60°. Let’s find the radius of
n
3
n
+ 3
n
+ 4
n
n
Is
n a
solution?
2
−1
1
3 + 3 + 4 = 10
1
0
= 1
the circle containing arc DE. To do so, we
→
10 ≠ 1
NO
2
12 + 6 + 4 = 22
2
1
= 2
→
22 ≠ 2
NO
drop a perpendicular from B to point F on
3
27 + 9 + 4 = 40
3
2
= 9
→
40 ≠ 9
NO
side AC.
4
48 + 12 + 4 = 64
4
3
= 64
→
64 = 64
YES
n = 4 Ans.
26. How many different combinations of 3
We are told that AB = 2. Because ∆ABF is a
numbers can be selected from the set {1, 2, 3,
30-60-90 right triangle, we know that
4, 5, 6, 7, 8} so that the numbers could
AF = ½(2) = 1 and the circle containing are
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