CAREER POINT
RMO-2011 EXAMINATION
Regional Mathematical Olympiad -2011
Time : 3 Hours
December 04, 2011
Instructions :
• Calculators (In any form) and protractors are not allowed.
fdlh Hkh rjg ds dsYdqysVj ;k dks.kekid dk iz;ksx oftZr gSA
• Rulers and compasses are allowed.
:yj vkSj dEikl dk iz;ksx fd;k tk ldrk gSA
• Answer all the questions. Maximum marks 100.
lHkh iz'uksa ds mÙkj nsaA dqy fu/kkZfjr vad 100 gSaA
• Answer to each question should start on a new page. Clearly indicate the question number.
izR;sd u, iz'u dk mÙkj u, i`"B ls izkjaHk djsaA iz'u la[;k dk Li"V mYys[k djsaA
1.
Let ABC be a triangle. Let D, E, F be points respectively on the segments BC, CA, AB such that AD, BE, CF
concur at the point K. Suppose BD/DC = BF/FA and ∠ADB = ∠AFC. Prove that ∠ABE = ∠CAD.
ABC ,d f=kHkqt gSA ekukfd D, E, F Øe'k% [k.M BC, CA, AB ij bl rjg gS fd AD, BE, CF fcUnq K ij laxkeh gSaA
eku yhft, fd BD/DC = BF/FA rFkk ∠ADB = ∠AFC. fl) dhft, fd ∠ABE = ∠CAD.
[12]
Sol.
A
E
F 2
K
3 7
5
B
6
1
4
C
D
Q ∠ADB = ∠AFC
∴ ∠1 = ∠ 2 = x (let)
…(i)
∴ ∠3 = (180 – x) (linear property).
∴ ∠BFK + ∠KDB = 180°
∴ BDKF is cyclic
∴ ∠5 = ∠2 – ∠BKF
but ∠ 5 = ∠ 6
…(ii)
{in ∆BKF, exterior angle} …(iii)
(same segment)
∴ ∠5 = ∠ 6 = (∠2 – ∠ 7)
Also
…(iv)
BD BF
=
DC FA
∴ by Thalse theorem FD || AC
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CAREER POINT
RMO-2011 EXAMINATION
∴ ∠4 = ∠BCA = ∠DCA
(corresponding)
…(v)
Now in ∆ADC
∠CAD = ∠ADB – ∠C
= x – ∠C
(by (i))
= ∠4 + ∠ 6 – ∠C
∠CAD = ∠C + ∠2 – ∠7 – ∠C
(by eqn. (iv) & (v))
∴ ∠CAD = ∠ 2 – ∠ 7
…(vi)
∴ by equation (iv) and (vi)
∠5 = ∠CAD
⇒ ∠ABE = ∠CAD
2.
Proved.
Let (a1, a2, a3,……, a2011) be a permutation (that is a rearrangement) of the numbers 1, 2, 3,……., 2011.
Show that there exist two numbers j, k such that 1 ≤ j < k ≤ 2011 and |aj – j| = |ak – k|.
ekukfd (a1, a2, a3,……, a2011) 1, 2, 3,……., 2011 la[;kvksa dk ,d Øep; gS (vFkkZr~ iqufoZU;kl gS) fn[kkb;s fd nks
la[;k,a j, k bl rjg gS fd 1 ≤ j < k ≤ 2011 rFkk |aj – j| = |ak – k|.
Sol.
[19]
Let j, k taken all x axis and rearrangement (a1, a2, a3,……, a2011) be on y-axis.
y
α
ak = aj + α
α
aj's
ak = a j – α
O
1
2
3 4
k
j
x
2011
α
Let k – j = α
Hence for any j & k at diff. of 'α'
Two points on y-axis can be taken on difference of 'α'.
3.
A natural number n is chosen strictly between two consecutive perfect squares. The smaller of these two
squares is obtained by subtracting k from n and the larger one is obtained by adding l to n. Prove that n – kl
is a perfect square.
,d izkÑfrd la[;k n nks Øekxr iw.kZoxks± ds chp esa pquh xbZ gSA bu nks oxks± esa NksVh okyh la[;k n esa ls k dks ?kVkus
ij izkIr gksrh gS rFkk cM+h okyh la[;k n esa l dks tksM+us ij izkIr gksrh gSA fl) dhft, fd n – kl ,d iw.kZ oxZ gSA
[12]
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CAREER POINT
RMO-2011 EXAMINATION
Sol.
Let λ2 and (λ + 1)2 are the consecutive perfect squares.
So n – k = λ2
…(i)
2
n + l = (λ +1)
…(ii)
Now, n – kl
= λ2 + k – k [(λ + 1)2 – n] [Using (i) & (ii)]
= λ2 + k – k [(λ + 1)2 – λ2 – k]
= λ2 + k – k [1 + 2λ – k]
= λ2 – 2λk + k2 = (λ – k)2
which is a perfect square
4.
Consider a 20-sided convex polygon K, with vertices A1, A2, ...., A20 in that order. Find the number of ways in
which three sides of K can be chosen so that every pair among them has at least two sides of K between
them. (For example (A1A2, A4A5, A11A12) is an admissible triple while (A1A2, A4A5, A19A20) is not.)
[19]
ekuk fd K ,d 20-Hkqtk okyk voeq[k cgqHkqt gS ftlds 'kh"kZ A1, A2, ...., A20 mlh Øe esa gSA Kkr dhft, fd ,sls
fdrus rjhds gSa ftlesa K dh rhu Hkqtk,¡ bl rjg pquh tk ldrh gSa fd muesa ls izR;sd ;qXe ds chp esa K dh de ls
de nks Hkqtk,a gksaA (mnkgj.kkFkZ (A1A2, A4A5, A11A12) xzkg; f=kd gS tcfd (A1A2, A4A5, A19A20) ugha gS)
Sol.
Total ways to select 3 sides = 20C3
Now, No. of ways of selection when exactly three sides are common = 20
i.e. [(A1A2, A2A3,A3A4), (A2A3, A3A4, A4A5)] etc.
A19
A20 A1 A2
A3
A4
A5
Now, No. of ways when exactly two sides are common.
No. of selection = 20 × 16
[If we select (A1A2, A2A3) then third cannot be selected from A20A1 & A3A4]
Now no. of ways of selection when the selected two sides are let A1A2 & A3A4 (i.e. a gap of 1 side) = 20
A19
A20 A1 A2
A3
A4
A5
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CAREER POINT
RMO-2011 EXAMINATION
Here two cases arise for third side
(i) Third side selected at a gap of two from above selected two sides
ways of selection of third side = 13
∴ Required ways = 20 × 13
(ii) Third side is at a gap of one side from any of above selected side
∴ Total ways = 20
Hence total ways of selection = 20C3 – 20 – 20 × 16 – 20 × 13 – 20 = 520
5.
Let ABC be a triangle and let BB1, CC1 be respectively the bisectors of ∠B, ∠C with B1 on AC and C1 on AB.
Let E, F be the feet of perpendiculars drawn from A onto BB1, CC1 respectively. Suppose D is the point at
which the incircle of ABC touches AB. Prove that AD = EF.
[19]
ekuk fd ABC ,d f=kHkqt gS rFkk BB1, CC1 Øe'k% ∠B, ∠C ds n~foHkktd gS] AC ij fcUnq B1 gS rFkk AB ij fcUnq C1
gSA ekuk fd E, F yEc ds pj.k gSa tks fd Øe'k% A ls BB1, CC1 ij [khaps x, gSaA eku yhft, fd D og fcUnq gS ftl
ij ABC dk var% oÙ̀k AB dks Li'kZ djrk gSA fl) dhft, AD = EF
A
C1
π−
F
Sol.
B+C
2
I
B1
E
B
AF = b sin
C
2
AE = c sin
B
2
∠FAE =
C
B+C
2
By cosine rule
 B+C 
(EF)2 = (AF)2 + (AE)2 – 2(AF) (AE) cos 

 2 
= b2 sin2
C
B
A
B
C
+ c2 sin2
– 2 bc sin sin sin
2
2
2
2
2
= (s – a)2
EF = (s – a)
AD = r cot
A
2
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CAREER POINT
RMO-2011 EXAMINATION
= 4R sin
A
B
C
A
sin sin cot
2
2
2
2
= (s – a)
AD = (s – a)
Hence AD = EF
6.
Find all pairs (x, y) of real numbers such that 16 x
2
+y
2
+ 16 x + y = 1
okLrfod la[;kvksa ds lHkh ;qXe (x, y) bl rjg Kkr dhft, tgk¡ 16 x
Sol.
16 x
2
+y
2
+y
[19]
2
+ 16 x + y = 1
2
+ 16 x + y = 1
Q A.M ≥ G.M
16 x
2
+y
+ 16 x + y
2
2
≥ 16 x
2
+ x+ y2 + y
2
2
1
≥ 4 x + x+ y + y
2
2–1 ≥ 2 2( x
2
+ x+ y2 + y)
x2 + x + y2 + y ≤ –
1
2
2
2
1 
1
1
1
1

+ y+  –
≤–
x +  –
4
4
2
2
2




2
2
1
1


x +  + y +  ≤ 0
2
2


True if
x=–
1
1
; y=–
2
2
 1 1
Ordered pair is  − ,− 
 2 2
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