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69 At Right Angles | Vol. 2, No. 2, July 2013 Vol. 2, No. 2, July 2013 | At Right Angles 69
Problems for the
Senior School
Problem editors: PRITHWIJIT DE & SHAILESH SHIRALI
Problems for Solution
Problem II-2-S.1
A circle has two parallel chords of length xx that are xx
units apart. If the part of the circle included between
the chords has area 2+ππ, �ind the value of xx.
Problem II-2-S.2
The prime numbers pp and qq are such that pp ppp
and pp p ppp are both perfect squares. Determine
the value of pp.
Problem II-2-S.3
Determine the value of the in�inite series
1
3� + 1 +
1
4� + 2 +
1
5� + 3 +
1
6� + 4 +⋯.
Problem II-2-S.4
In trapezium AAAAAAAA, the sides AAAA and BBBB
are parallel to each other; AAAA A A, BBBBBB, CCCCC C,
AAAA A AA. Sides AAAA and CCCC are
extended to meet at EE. Determine the magnitude
of ∡AAAAAA.
Problem II-2-S.5
You are told that the number 27000001 has
exactly four prime factors. Find their sum.
(Computer solution not acceptable!)
Solutions of Problems in Issue-II-1
Solution to problem I-2-S.1 Drawn through the
point AA of a common chord AAAA of two circles is a
straight line intersecting the �irst circle at the point
CC, and the second circle at the point DD. The tangent
to the �irst circle at the point CC and the tangent to
the second circle at the point DD intersect at the
point MM. Prove that the points MM, CC, BB, and DD are
concyclic. (See Figure 1.)
Two cases are possible: (i) CC and DD are on the
same side of the line joining the centres of the
circle. (ii) CC and DD are on the opposite sides of the
line joining the centres of the circle. In both cases
we see that ∡MMMMMMM ∡CCCCCC and ∡MMMMMMM ∡AAAAAA.
Thus ∡CCCCCCC ∡CCCCCCC ∡AAAAAA A
∡MMMMMMM ∡MMMMMMMMMM∘ − ∡CCCCCC. Therefore
points MM, CC, BB, and DD are concyclic.
Solution to problem I-2-S.2 In triangle AAAAAA,
point EE is the midpoint of the side AAAA, and point DD is
the foot of the altitude CCCC. Prove that ∡AA A A∡BB if
and only if AAAA A AAAAA. (See Figure 2.)
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70 At Right Angles | Vol. 2, No. 2, July 2013 Vol. 2, No. 2, July 2013 | At Right Angles 70
A
B
C
D
M
Figure 1. Showing that M, C, B, D are concyclic
(Problem I-2-S.1)
A
B C
E
D
F
Figure 2. Showing that ∡A = 2∡B if and only if
AC = 2ED (Problem I-2-S.2)
Let FF be the midpoint of BBBB. Therefore EEEE E EEEE,
and AAAAA AAAAA. Also in right-angled △CCCCCC, FF is the
midpoint of the hypotenuse BBBB. Therefore
CCCCC CCCCCCCCC.
In △BBBBBB, DDDDDDDDD. So ∡FFFFFFF ∡FFFFFFF ∡BB.
Since EEEE E EEEE, ∡FFFFFFF ∡AA. But
∡FFFFFFF ∡FFFFFF F ∡DDDDDD. That is,
∡AAA ∡BB B ∡DDDDDD. Now
∡AAA A∡BB B ∡DDDDDDD ∡BB B BBBB
= EEEEE EEEEEEEEEEEEEEEEE
Solution to problem I-2-S.3 Solve the
simultaneous equations: aaaa a aa a aa a a,
bbbbbbbbbbb b, cccccccccccc, ddddddddddd d,
where aa, bb, cc, dd are real numbers.
Adding the four equations we obtain
(aa a aaaaaa a aaa aaaaa a aaa aaaaa a aaa a aaa (1)
Adding the �irst two equations we obtain
(bbbbbbbbbbbbbbbbb bb (2)
Adding the last two equations gives
(ddddddddddddddddddd (3)
Subtracting (3) from (2) yields
(aa a aa a aaaaa a aaa a a. Thus either aa a aa aa or
bbbbb. But, if bbbbb then bbbbbbbbbbbbbbbbbbbbb
which leads to 5 = 2, an absurdity. Therefore
aa a aa aa. Now from (1) we get bbbbbbb. So
ccccccccccccccc. Therefore:
3 = aaaa a aa a aa a aaaa aa a aaa a aaaa (4)
2 = ccccccccccccccccccccccccccccc
= 6+ aaaa a aaaaaaa (5)
These lead to:
aa a aa a aaaa aaa (6)
2aa a aa a aaaa a aa (7)
From (6) and (7), aa aa. Hence ccccccccc.
Therefore bbbbbbbbbbbbbb and ddddddddd. It
is easy to see that these values satisfy the given
equations. Therefore (aaa aaa aaa aaa aaaa aa aa aa.
Solution to problem I-2-S.4 Let xx, yy, aa be positive
numbers such that xx� + yy� = aa. Determine the
minimum possible value of xx� + yy� in terms of aa.
We have: xx�+yy� = (xx�+yy�){(xx�+yy�)�−3xx�yy�} =
aaaaa� −3xx�yy�). Hence xx� + yy� attains its minimum
value when xx�yy� attains its maximum value. Since
xx� + yy� = aa, the maximum possible value of xx�yy�
is (aaaaa� = aa�/4, with equality attained only when
xx xxxx �aaaa. Hence the minimum value of
xx� + yy� is aa �aa� −3aa�aa� = aa�/4.
Solution to problem I-2-S.5 Let pp, qq and yy be
positive integers such that pp is greater than qq, and
yy� − qqqqqqqqqqq. Prove that pp� − qq� is not a
prime number.
Suppose pp�−qq� = (ppppppppppppp is a prime number.
Then pp ppppp and therefore yy� − qqqqq qq qq. So
qq qqq�/(yy yyyy yy yyyyyyyy yyy.
As qq and yy are integers, so is 1/(yy yyy. Since
yy y y, it must be that yy yy. Hence qq qq and
pp ppppppp, giving pp� − qq� = 9, which is not
prime. A contradiction. Therefore pp� − qq� is not a
prime number.
2 At Right Angles ∣ Vol. 1, No. 3, June 2013