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JMO mentoring scheme answers
Charitable Trust
January 2012 paper
1
Ans : 1295
The sum of the first n integers is ½n(n + 1). Can you prove this, e.g. by drawing a square pattern of dots?
2
Ans : 1
Since 8n + 50 = 4(2n + 1) + 46, we require 2n + 1 to divide into 46 whose factors are 1, 2, 23 and 46.
Since 2n + 1 can not equal 1, 2 or 46, then 2n + 1 = 23.
The answer depends on the number of usable factors of 46 which is the remainder after dividing 8n + 50 by 2n + 1.
3
Ans : (i) ½d²
(ii) (3 + 2Ö 2)d²
If the octagon is partitioned as suggested in the hints, the sides of the rectangles are d and x, say.
Then we have x ² + x ² = d ² so x = 1 ¸ Ö 2.
4
Ans : 2 : 5
One easy way to proceed is to rotate the whole figure through 180° about F. We can now see that MN is one of five equal
segments along the line AA´ where A´ is the image of A after the rotation. F is the mid-point of AA´.
5
Ans : 1 011 030
There are ½ ´ 1006 ´ 1005 ways of choosing pairs of odd numbers, e.g. (1, 3), (1, 5), …, (57, 2003), … (2009, 2011).
There are the same number of ways of choosing pairs of even numbers so the answer is 1006 ´ 1005.
6
Ans : 1 : 27
It is easiest to work backwards when drawing this. Start with the larger tetrahedron then draw ABCD inside it. A, B, C and
D must lie at the ‘middle’ of the triangular faces of the larger tetrahedron. By referring to note F on medians of triangles,
you will be able to deduce that the length of the edges of ABCD must be ⅓ of the lengths of the edges of the larger
tetrahedron. To convince yourself of this, draw out the cross-section of one of the reflection planes of the tetrahedrons.
7
(i) If a and b are both even, then c is even and there will be a common factor. If a and b are both odd, then c ² is even but it
is not a multiple of 4 which must be the case if c is even. Hence one of a and b is even, the other odd. This makes c odd.
(ii) Use (m ² - n ²)(m ² - n ²) = m 4 - 2 m ²n ² + n 4, (2 mn) ² = 4m ²n ², (m ² + n ²)(m ² + n ²) = m 4 + 2 m ²n ² + n 4.
8
Ans : k = 6, 1 ± Ö 15 or -1 ± Ö 39.
We note that AB = Ö 40.
(a) For CA = CB, C would lie on y = 3x through (0, 0) and (1, 3) which is the perpendicular bisector of AB.
Since x = 2 then y = 6.
In all other cases the point C must lie on a circle with radius Ö 40 with centre either at A (AC = AB) or at B (BC = BA).
(b) x = 2 meets the circle centre A at two possible positions of C, one below (2, 1) and one above.
Let the mid-point of the chord between these positions be Q.
AQ = 5 and AC = Ö 40 so by Pythagoras’s rule, CQ = Ö 15. Hence k = 1 + Ö 15 or 1 - Ö 15.
(c) x = 2 meets the circle centre B at two possible positions of C, one below (2, -1) and one above.
Let the mid-point of the chord between these positions be R.
BR = 1 and BC = Ö 40 so by Pythagoras’s rule, CR = Ö 39. Hence k = -1 + Ö 39 or -1 - Ö 39.
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See UKMT website www.ukmt.org.uk > Mentoring > Junior > Ideas in .. for notes B and F.
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