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JMO mentoring scheme answers
July 2012 paper
1
Charitable Trust
Ans : 18
To be divisible by 9, the digits must add to a multiple of 9. The answer must be even to be divisible by 11.
2
Ans : 6 pairs
Find the factors pairs of 105 and 150. You can not use all the factor pairs for 150.
3
There are 18 sums. Consider trying to prove the opposite of what is required, i.e. that there are not more than 2 of each
sum. Since there are nine sums lying between 0 and 8, each sum would have to appear twice. (A)
A sum of 8 can not appear in a diagonal otherwise it is impossible to obtain sum 0. Suppose the sum 8 appears in a row;
then there is another row with sum 8 (by statement A). Sum 0 must then appear in two other rows. Now no diagonal nor
column can have sum 1 or sum 7 so these sums must account for the remaining four rows.
In the four rows with sum 0 or sum 1, there are just two 1s altogether (the other 30 being 0s). Similarly in the four rows
with sum 7 or sum 8, there are just two 0s. These four digits can be placed in a maximum of four columns. Hence there are
at least four columns with four 1s and four 0s, i.e. having sum 4, contradicting our assumption.
4
ÐABD = ÐADB because ABCD is a rhombus. Hence ÐPBD = ÐPDB which makes D PBD isosceles.
Now we have PB = PD, AB = AD (rhombus) and ÐABD = ÐADB so D PAB is congruent to D PAD.
This proves that ÐPAB = ÐPAD. Since ÐCAB = ÐCAD and the four angles add to 360°, then ÐPAB + ÐCAB = 180°.
5
Let the middle number be x.
(x - 1)³ = (x - 1)(x - 1)² = (x - 1)(x² - 2x + 1) = x³ - 2x² + x - x² + 2x - 1 ;
(x + 1)³ = (x + 1)(x + 1)² = (x + 1)(x² + 2x + 1) = x³ + 2x² + x + x² + 2x + 1 .
Adding these plus the middle x³ gives 3x³ + 6x and the sum of the three original numbers is 3x.
6
Ans : 9
If x is the number of tied matches, the points gained are 3(13 - 2x) + x + 0. If this is 29, then 39 - 5x = 29 and x = 2.
7
We work out that (a + b)² + (a - b)² = 2a² + 2b² .
Since c is the hypotenuse, we can write (a + b)² + (a - b)² = 2c² .
Every square number is 0 or positive so (a + b)² ≤ 2c² .
8
(Ö 5 + 1)(Ö 5 - 1) = 5 + Ö 5 - Ö 5 - 1 = 4.
(a)
Since SM bisects ÐAMO, OS : SA = OM : MA follows from the result of (7).
This means that OS : OS + SA = OM : OM + MA.
Take OA = 2 ; this saves getting fractions in the calculations. MA = Ö 5 using Pythagoras’ rule.
Hence OS : OA = 1 : 1 + Ö 5 .
This ratio is also Ö 5 - 1 : 4 by multiplying up the ratio by Ö 5 - 1.
(b)
[NB a : b = ka : kb, eg. 2 : 7 = 6 : 21].
(i)
Since BT = BO and BS is perpendicular to OT, S is the mid-point of OT. Hence OT = 2 ´ OS = ½(Ö 5 - 1).
(ii)
RT = 1 + ½(Ö 5 - 1) = 1 + ½Ö 5 - ½ = ½(Ö 5 + 1) .
(iii) BT : OT = 1 : ½(Ö 5 - 1) and RT : BT = ½(Ö 5 + 1) : 1 = 1 : ½(Ö 5 - 1) by multiplying the ratio by ½(Ö 5 - 1).
(iv) Consider D OTB and D BTR. ÐOTB is common to both triangles. With the ratios in (iii) proved equal, the triangles
must be similar. Since D OTB is isosceles, then D BTR is isosceles. Let ÐOTB = x, then ÐOBT = 180° - 2x.
ÐBRT = 180° - 2x by the similar triangles and ÐOBR = 180° - 2x because OR = OB.
The angles round B add to 360° - 4x . Because D BTR is isosceles, this is the same as ÐOTB .
360° - 4x = x hence x = 72°.
This yields the angles as required.
This solution uses ratios rather than fractions. It is worth the effort learning to understand and manipulate ratios; it can remove the
problems of deciding on units or working with fractions within fractions. The initial choice to make OA = 2 helped find MA easily.
For those who know a little trigonometry, we have proved that cos 72° = ¼(Ö 5 - 1).
We have also shown that the golden ratio ½(Ö 5 + 1) : 1 is closely connected to the regular pentagon.
½(Ö 5 - 1) is the reciprocal value of ½(Ö 5 + 1). The Greek letter ‘phi’ written ϕ is often used to represent ½(Ö 5 + 1) ≈ 1.618003…