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problem corner

Vol. 3, No. 1, March 2014 | At Right Angles 71

Problems for the

Senior School

Problem Editors: Prithwijit De & Shailesh Shirali

Problems for Solution

Solutions of Problems in Issue-II-3

Problems for Solution

Problem III-1-S.1

Let ffffffffffff ff ffffffff� + bbbbffff f bbbb, where ffffaa bbbbaa bbbb are positive

integers. Show that there exists an integer mmmm such

that ffffffmmmmff is a composite number.

Problem III-1-S.2

Show that the arithmetic progression

1aa 5aa 9aa 13aa 17aa 21aa 25aa 29aa ... contains in�initely

many prime numbers.

Problem III-1-S.3

In △AAAAAAAAAAAA, the midpoint of AB is DDDD, and EEEE is the

point of trisection of AAAAAAAA closer to AAAA. Given that

∡AAAADDDDAAAA ff fAAAAAAAAEEEE, determine the magnitude of

∡AAAAAAAAAAAA.

Problem III-1-S.4

We know that a median of a triangle bisects it into

two triangles of equal area. We also know that the

medians of a triangle are concurrent. Given a

△AAAAAAAAAAAA, does there necessarily exist a point DDDD on

side AAAAAAAA such that △AAAAAAAADDDD and △AAAAAAAADDDD have equal

perimeter?

If such a point exists, then we can similarly obtain

points EEEE and FFFF on AAAAAAAA and AAAAAAAA, respectively such

that AAAAAAAA and AAAAAAAA bisect the perimeter of AAAAAAAAAAAA. Are

the lines AAAADDDDDD AAAAEEEEEE AAAAFFFF concurrent?

Problem III-1-S.5

Let AAAA ff f���� and AAAA AA BB����. Is BB� + 5� a prime

number? Justify your answer.

Solutions of Problems in Issue-II-3

Solution to problem II-3-S.1

Let PPPP be a polynomial such that

PPPPffffffff ff PPPPffPPff f PPPPfffffffff f PPPPfffffffff� and PPPPffPP1ff ff f.

Find PPPPfffff.

Putting ffff ff f yields PPPPffPPff f PPPPfffff ff PP. Putting

ffff ff fff, we get:

PPPPffPP1ff ff PPPPffPPff PP PPPPfffff f PPPPfffffff

hence PPPPfffff ff PP1. Putting ffff ff f, we get:

PPPPfffff ff PPPPffPPffPP2PPBBPPPPffPPff ff PP2PP3PPPPffPPff ff PP2+3PPPPfffffff

hence PPPPfffff ff f and PPPPffPPff ff PP1. Therefore

PPPPffffffff ff PP1 PP ffff f ffff� and PPPPfffff ff f.

Solution to problem II-3-S.2

In △AAAAAAAAAAAA, the midpoint of AAAAAAAA is DDDD; the foot of the

perpendicular from AAAA to AAAAAAAA is EEEE; the foot of the

perpendicular from DDDD to AAAAAAAA is FFFF; AAAAAAAAAAA, EEEEAAAA AAA;

area of △AAAAAAAAAAAA is 8BB. Find EEEEFFFF.

There are two cases: (i) EEEE lies between AAAA and AAAA.

(ii) EEEE lies to the left of AAAA on the line AAAAAAAA. Possibility

(i) is depicted in Figure 1.

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Problems for Solution

Problem III-1-S.1

Let ffffffffffff ff ffffffff� + bbbbffff f bbbb, where ffffaa bbbbaa bbbb are positive

integers. Show that there exists an integer mmmm such

that ffffffmmmmff is a composite number.

Problem III-1-S.2

Show that the arithmetic progression

1aa 5aa 9aa 13aa 17aa 21aa 25aa 29aa ... contains in�initely

many prime numbers.

Problem III-1-S.3

In △AAAAAAAAAAAA, the midpoint of AB is DDDD, and EEEE is the

point of trisection of AAAAAAAA closer to AAAA. Given that

∡AAAADDDDAAAA ff fAAAAAAAAEEEE, determine the magnitude of

∡AAAAAAAAAAAA.

Problem III-1-S.4

We know that a median of a triangle bisects it into

two triangles of equal area. We also know that the

medians of a triangle are concurrent. Given a

△AAAAAAAAAAAA, does there necessarily exist a point DDDD on

side AAAAAAAA such that △AAAAAAAADDDD and △AAAAAAAADDDD have equal

perimeter?

If such a point exists, then we can similarly obtain

points EEEE and FFFF on AAAAAAAA and AAAAAAAA, respectively such

that AAAAAAAA and AAAAAAAA bisect the perimeter of AAAAAAAAAAAA. Are

the lines AAAADDDDDD AAAAEEEEEE AAAAFFFF concurrent?

Problem III-1-S.5

Let AAAA ff f���� and AAAA AA BB����. Is BB� + 5� a prime

number? Justify your answer.

Solutions of Problems in Issue-II-3

Solution to problem II-3-S.1

Let PPPP be a polynomial such that

PPPPffffffff ff PPPPffPPff f PPPPfffffffff f PPPPfffffffff� and PPPPffPP1ff ff f.

Find PPPPfffff.

Putting ffff ff f yields PPPPffPPff f PPPPfffff ff PP. Putting

ffff ff fff, we get:

PPPPffPP1ff ff PPPPffPPff PP PPPPfffff f PPPPfffffff

hence PPPPfffff ff PP1. Putting ffff ff f, we get:

PPPPfffff ff PPPPffPPffPP2PPBBPPPPffPPff ff PP2PP3PPPPffPPff ff PP2+3PPPPfffffff

hence PPPPfffff ff f and PPPPffPPff ff PP1. Therefore

PPPPffffffff ff PP1 PP ffff f ffff� and PPPPfffff ff f.

Solution to problem II-3-S.2

In △AAAAAAAAAAAA, the midpoint of AAAAAAAA is DDDD; the foot of the

perpendicular from AAAA to AAAAAAAA is EEEE; the foot of the

perpendicular from DDDD to AAAAAAAA is FFFF; AAAAAAAAAAA, EEEEAAAA AAA;

area of △AAAAAAAAAAAA is 8BB. Find EEEEFFFF.

There are two cases: (i) EEEE lies between AAAA and AAAA.

(ii) EEEE lies to the left of AAAA on the line AAAAAAAA. Possibility

(i) is depicted in Figure 1.

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72 At Right Angles | Vol. 3, No. 1, March 2014

(i) We have [AAAAAAAAAAAAAA AA AAAA AA AAAAAA AA AAAA AA AAAAAAAA, so

AAAAAAAA AA AAAA, AAAAEEEE AA AAAAAAAA EE EEEEAAAA AA EE EE EE AA AA. Now assign

coordinates: AAAA AA EEEEEE EEEE, AAAA AA EEEEEE AAAAEE, AAAA AA AAAABBEE EEEE,

AAAA AA AAAAAAAAAA, EEEE EE EEEEEE EEEE.

The slope of AAAAAAAA is EEEEEE3, and the slope of EEEEDDDD is

3AAAA. The equations of AAAAAAAA and EEEEDDDD are

xxxxAAAA xx xxxxAAAAAA AA AA and xxxx xxxxxxxxxxxxxxxxx xxxxxx,

respectively. Solving these two equations for xxxxEE xxxx,

we get DDDD DDDDDDFFAAAAAAAAAA AAAAAAAAAAAA. Hence

AAAAEEEE AA �EEEEEEEE� xx xxxx�EEEEEEEEEE√3EEEEEE.

(ii) Following the same steps we get AAAA AA EEEEEE EEEE,

AAAA AA AABBEE EEEE, AAAA AA AAAAAAAAAA, EEEE EE EEEEEE EEEE. The equations of

AAAAAAAA and EEEEDDDD are now xxxxAAAA xx xxxxAAAAAA AA AA and

xxxx xxxxxxxxxxxxxxxxxxx xxxxxx. Solving these we get

DDDD DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD, so AAAAAAAA AA AAAA√EEEEEEEEEEEEEEEE.

Solution to problem II-3-S.3

In how many ways can the numbers EEEE, EEEE, EEEE, ...,

FF, EE, AA be arranged in a line so that the absolute

values of the numbers do not decrease from left to

right?

If the stated property is to be satis�ied, then for

each aaaa aa aaAAAA AAAA aa EE EEaa, the numbers aaaa, EEaaaa must

occur together. Further, as the absolute values of

the entries must not decrease, the numbers

AAAA AAAAAAAA AA must appear in this order, reading from

left to right. If these two conditions are met, then

the stated property will hold good. Having �ixed

such an arrangement, we observe that for each aaaa,

the number EEaaaa must occur either immediately to

the left or immediately to the right of aaaa. Thus

there are just two ways to insert EEaaaa for any aaaa.

Hence the total number of legitimate

arrangements is AA� AA AAAAAA.

Solution to problem II-3-S.4

Two ships sail with constant speed and direction. It

is known that at 9:00 am the distance between

them was AAAA miles; at 9:35 am, AAAA miles; and at

9:55 am, AA3 miles. What was the least distance

between the ships, and at what time was it

achieved? [IMO Short list, 1968]

As the ships are sailing with constant speed and

direction, the second ship is sailing at a constant

speed and direction with reference to the �irst

ship. Let AAAA be the constant position of the �irst

ship in this frame, and let l be the path of the

second ship in relation to the �irst one. Let points

AAAA�, AAAA�, AAAA�, and AAAA on l be positions of the second

ship with respect to the �irst ship at 9:00, 9:�5,

9:55, and the moment the two ships are closest to

each other. Then we have the following equations:

AAAAAAAA� AA AAAAAA AAAAAAAA� AA AAAAAA AAAAAAAA� AA AA3;

AAAA�AAAA� ∶ AAAA�AAAA� AA AA ∶ AA, AAAAAAAA�

� AA AAAAAAAA� xx xxxxxxxx�

� .

We now adopt coordinates (see Figure 2). Let l be

taken to be the xxxx-axis, with AAAA� AA AAAAAAAAAA,

AAAA� AA AAAAccccEE EEEE, AAAA� AA AAAAAAccccEE EEEE where cccc cc EE; it is

assumed that the second ship is moving along the

xxxx-axis in the positive direction; let AAAA AA EEEEEEEE AAAAEE.

Then we have:

aaaa� xx xxxx� AA AAAA�EE

EEaaaa EE EEEEEEEE� xx xxxx� AA AAAA�EE

EEaaaa EE EEEEEEEEEE� xx xxxx� AA AA3�.

These yield, by subtraction:

AAAAaaaacccc cccccccccc� AA AAAA� EE EEEE� AA AAAAAAAA

AAAAaaaacccc cccccccccccc� AA AAAA� EE EE3� AA AA3AA.

Treating these as a pair of simultaneous equations

in aaaacccc and cccc� we get aaaacccc cccccc, cccc� AA AA, hence cccc AA AA,

aaaa AA AAAA. This in turn yields AAAA AA bbAAAA. Hence the

closest distance of AAAA from l is AAAA miles, and the

time at which this takes place is AAAA AA AA minutes

after 9:00, i.e., at 10:20 am.

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Figure 1.

Figure 2.