Download Solutions

For more information about
the mentoring schemes, and
how to join, visit :
www.mentoring.ukmt.org.uk
JMO mentoring scheme answers
June 2014 paper
1
Ans : x = -¾
The x2 s cancel out to give the equation -3 = 4x.
2
Ans : 1007
In a group of 10 numbers whose units digits run between 0 and 9, the number of even digits alternates between even and
odd as the you go through the list. Thus half of them have the required property. This accounts for the numbers between 10
and 2009 of which there are 1000. The other ones are 2, 4, 6, 8, 2010, 2012 and 2014.
3
Ans : (a) about 8 m (b) about 2.592 km from the South Pole
(a) Taking a side view of the earth we can see that RP ≈ NP = 10 km. Using Pythagoras’s rule
we obtain :
OR2 ≈ 63702 - 102 which you might note is equal to (6370 - 10)(6370 + 10) = 6360 ´ 6380.
From this we can calculate that OR is about 8 m short of 6370 km.
(b) One possible starting point P1 is near the South Pole. Travel 10 km towards the South Pole
to a point T where turning east and following a parallel of latitude round the South Pole takes
you a further 10 km back to T. Then travel north back to the starting point P1.
The radius of the parallel of latitude r is given by 2pr = 10 so r = 1.592 km.
The starting point P1 needs to be about 2.592 km from the South Pole.
There are further positions for P closer to the South Pole allowing you to circle the South Pole
two or more times.
4
Ans : May 17th when Emily is 8 years old
Let March 25th be x days before Emily's birthday.
Using the information for March 25th we can form an equation :
Jack’s age = 8 ´ Emily’s age
365(n2 - 1) - 6 - x = 8(365n - x)
which reduces to :
7x = 371 + 365(8n - n2)
From this n must be 8 otherwise the event can not happen in the same year and so x = 53.
Allowing for 6 further days in March and 30 days in April gives Emily’s birthday.
5
Without loss of generality let the square have side length 2.
Using Pythagoras’s rule, we find AM = Ö2.
Thus the area of the quarter-circle with centre M = ¼ ´ p ´ (Ö2)2 = ½p
Area of semi-circle + area of DADM = ½ ´ p ´ 12 + 1.
Hence the area of the crescent is 1 while the area of rectangle ABCD is 2.
This is surprising as the answer does not involve p.
The problem is called the Lune of Hippocrates :
by tradition Hippocrates (of Chios in Greece, d. about 410 BC) worked on this problem.
Try the link http://www-history.mcs.st-and.ac.uk/Biographies/Hippocrates.html.
6
Ans : 1706
As the LCM of 2 and 3 is 6, we can assert that the pattern of pawns will be repeated every block of 6 by 6 squares. We note
that if a coordinate is a multiple of 6, it can count as either the multiple or 3 or the even number. Such a block can take 11
pawns and there are 144 such blocks. The rest of the board can be split into 24 blocks of 3 by 6 squares which can take 5
pawns each and one block of 3 by 3 squares which can take 2 pawns.
7
Since AO = OD (given), using OP = OR (radii) and ÐOAB = ÐODC = 90°,
we can prove DODR is congruent to DOAP. Hence ÐOAB = ÐODC.
Since the lengths of tangents to a circle are equal,
i.e. BP = BQ and CQ = CR,
there are two further pairs of 90° congruent triangles.
In summary, DOBP ≡ DOBQ, DOCQ ≡ DOCR and DODR ≡ DOAP.
The sum of the angles around O is 180°.
Thus, for example, ÐAOP + ÐPOB + ÐCOR = 90° .
Hence ÐAOB = ÐOCR, i.e. ÐAOB = ÐOCD.
Thus DAOB is similar to DDCO. [Note the ordering of the letters.]
It follows that OA : CD = AB : OD
which leads to the required result.