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1.
Determine all 3 digit numbers N which are divisible by 11 and where N/11 is
equal to the sum of the squares of the digits of N.
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Solution Answer: 550, 803.
Personally, I hate this type of question. The fastest way to solve it
is almost certainly to scan the 81 multiples of 11 from 110 to 990.
However, to get much credit you have to adopt a more algebraic approach.
But a question which allows the enumeration approach is a bad question.
So, put N/11 = 10a + b. If a + b <= 9, we have 2a2 +
2ab + 2b2 = 10a + b, so b is odd and after a little more work we
find 550 is the only solution. If a + b > 9, we have (a+1)2 +
(a+b-10)2 + b2 = 10a + b, so b is odd and after a
little more work we find 803 is the only solution.
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2.
For what real values of x does the following inequality hold:
4x2/(1 - Ö(1 + 2x))2
< 2x + 9 ?
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Solution Answer: - 1/2 <= x < 45/8.
We require the first inequality to avoid imaginary numbers. Hence we
may set x = -1/2 + a2/2, where a >= 0. The inequality now
gives immediately a < 7/2 and hence x < 45/8. It is a matter of taste
whether to avoid x=0. I would allow it because the limit as x tends to 0 of
the lhs is 4, and the inequality holds.
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3.
In a given right triangle ABC, the hypoteneuse BC, length a, is divided into
n equal parts with n an odd integer. The central part subtends an angle a at A. h is
the perpendicular distance from A to BC. Prove that:
tan a = 4nh/(an2 -
a).
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Solution Let M be the midpoint of BC,
and P and Q the two points a/2n either side of it, with P nearer B. Then a = angle PAQ
= angle QAH - angle PAH (taking angles as negative if P (or Q) lies to the
left of H). So tan a = (QH - PH)/(AH2 + QH.PH) = AH.PQ/(AH2 +
(MH - a/2n)(MH + a/2n)) = (ah/n)/(a2/4 - a2/(4n2))
= 4nh/(an2 - a).
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4.
Construct a triangle ABC given the lengths of the altitudes from A and B and
the length of the median from A.
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Solution Let M be the midpoint of BC,
AH the altitude from A, and BI the altitude from B. Start by constructing
AHM. Take X on the circle diameter AM with MX = BI/2. Let the lines AX, HM
meet at C and take B so that BM = MC. [This works because CMX and CBI are
similar with MX = BI/2 and hence CM = CB/2.]
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5.
The cube ABCDA‘B’C‘D’ has A above A‘, B above B’ and so on. X is any point of
the face diagonal AC and Y is any point of B‘D’.
(a) find the locus of the midpoint of XY;
(b) find the locus of the point Z which lies one-third of the way along XY, so
that ZY=2XZ.
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Solution The key idea is that the
midpoint must lie in the plane half-way between ABCD and A‘B’C‘D’.
Similarly, Z must lie in the plane one-third of the way from ABCD to
A‘B’C‘D’.
(a) Regard ABCD as horizontal. Then the locus is the square with
vertices the midpoints of the vertical sides.
(b) The locus is a rectangle Ö2/3 x 2Ö2/3.
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6. A
cone of revolution has an inscribed sphere tangent to the base of the cone
(and to the sloping surface of the cone). A cylinder is circumscribed about
the sphere so that its base lies in the base of the cone. The volume of the
cone is V1 and the volume of the cylinder is V2.
(a) Prove that V1 ¹ V2;
(b) Find the smallest possible value of V1/V2.
For this case construct the half angle of the cone.
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Solution Let the radius of the
sphere be r and the half-angle of the cone q. Then the the cone‘s
height is (1 + 1/sin q), and the radius of its base is r(1 + 1/sin q) tan q. We
easily find V1/V2 = (1 + s)2/(s - s2),
where s = sin q. So V1/V2 = 1 requires 2s2 +
s + 1 = 0, which is impossible since 0 <= s < 1. Differentiating,
we find the minimum is at s = 1/3.
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