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Difference between revisions of "2012 IMO Problems/Problem 4"

(Created page with "First we construct three circles:the circumcircles of ABC() ,The circle() with centre A and radius AC and the circle( ) with centre B and radius BC. Note that the centre of lies...")
 
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First we construct three circles:the circumcircles of ABC() ,The circle() with centre A and radius AC and the circle( ) with centre B and radius BC.
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Find all functions <math>f: \mathbb{Z} \to \mathbb{Z}</math>  such that, for all integers <math>a, b,</math> and <math>c</math> that satisfy <math>a +  b+ c = 0</math>, the following equality holds:
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Note that the centre of lies on the midpoint of AB,so the three circles are co-axial with radical axis CD.
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<cmath>f(a)^2 + f(b)^2 + f(c)^2 = 2f(a)f(b) + 2f(b)f(c) + 2f(c)f(a).</cmath>
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Let AX=Y=ABX=Z=BAYBZ=P .
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(Here <math>\mathbb{Z}</math> denotes the set of integers.)
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so AYB=AZB=90,so X must be the orthocenter of ABP implying P lies on the radical axis.
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let us denote PWR(P) by the power of P wrt .[which are equal]
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==Solution==
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From similar triangles ABCACD  we get AC2=ADAB=AL2ALD=ABL
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Consider <math>a = b = c = 0.</math> Then <math>f(0)^2 + f(0)^2 + f(0)^2 = 2f(0)f(0) + 2f(0)f(0) + 2f(0)f(0) \Rightarrow 3f(0)^2 = 6f(0)^2 \Rightarrow</math> <cmath>f(0) = 0.</cmath>
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similarly BKD=BAK
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Now we look at <math>b = -a, c = 0.</math> <math>f(a)^2 + f(-a)^2 + f(0)^2 = 2f(a)f(-a) + 2f(-a)f(0) + 2f(0)f(a) \Rightarrow</math> <math>f(a)^2 + f(-a)^2 = 2f(a)f(-a) \Rightarrow</math> <math>f(a)^2 - 2f(a)f(-a) + f(-a)^2 = 0 \Rightarrow</math> <math>(f(a) - f(-a))^2=0 \Rightarrow</math> <cmath>f(a)=f(-a).</cmath>
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now APD=ABZ=ALZ implying ADLP is cyclic.
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so ALPL.....(1) meaning PL2=PWR(P).
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We can write <math>f(c)^2 - 2f(c)(f(a)+f(b)) + (f(a)-f(b))^2 = 0  \Rightarrow</math>
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similarly we get BKPK.....(2) meaning PK2=PWR(P)
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hence PK=PL.....(3)
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<cmath>f(c) = f(-c) = f(a+b) =\frac{2(f(a)+f(b)) \pm \sqrt{4(f(a)+f(b))^2 - 4(f(a)-f(b))^2}}{2}</cmath>
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so MKP=MLP[using (1),(2),(3)]
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proving MK=ML
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<cmath>\Rightarrow f(a+b) = f(a) + f(b) \pm 2\sqrt{f(a)f(b)}</cmath>
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If <math>f(b) = 0</math>, then <cmath>f(a+b) = f(a) = f((a)mod(b))</cmath>
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'''Case 1''': <math>f(1) = 0 \Rightarrow f(x)= 0</math> <math>\forall</math> <math>x.</math> <math>\Box</math>
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Case 2: <math>f(1) \not= 0</math>, we will have <math>f(2) = f(1) + f(1) \pm 2\sqrt{f(1)f(1)} \Rightarrow f(2) = 0</math> or <math>f(2) = 4f(1)</math>
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'''Case 2.1''': <math>f(1) \not= 0, f(2) = 0 \Rightarrow f(x) = f(x</math> <math>mod</math> <math>2) \Rightarrow f(x) = f(1)</math> if <math>x</math> is odd, <math>f(x) = 0</math> if <math>x</math> is even. <math>\Box</math>
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Case 2.2: <math>f(1) \not= 0, f(2) = 4f(1) \Rightarrow f(3) = f(2) + f(1) \pm 2\sqrt{f(2)f(1)}</math>
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<math> \Rightarrow f(3) = 5f(1) \pm 4f(1) \Rightarrow f(3) = f(1)</math> or <math>9f(1)</math>
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'''Case 2.2.1''': <math>f(1) \not= 0, f(2) = 4f(1), f(3) = f(1).</math>
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<math>\Rightarrow f(4) = f(1) + f(3) \pm 2\sqrt{f(1)f(3)}</math> and <math>f(4) = f(2) + f(2) \pm 2\sqrt{f(2)f(2)}</math>
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<math>\Rightarrow f(4) = f(1)</math> or <math>0</math> and <math>f(4) = 16f(1)</math> or <math>0</math>
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<math>\Rightarrow f(4) = 0 \Rightarrow  f(x) = f(x</math> <math>mod</math> <math>4).\Box</math>
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'''Case 2.2.2''': <math>f(1) \not= 0, f(2) = 4f(1), f(3) = 9f(1).</math>
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<math>\Rightarrow f(4) = f(1 + 3) =  f(1) + f(3) \pm 2\sqrt{f(1)f(3)} = 16f(1)</math> or <math>4f(1)</math>
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and <math>f(4) = f(2) + f(2) \pm 2\sqrt{f(2)f(2)} = 16f(1)</math> or <math>0. \Rightarrow f(4) = 16f(1).</math>
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If <math>x \le 4</math> then <math>f(x) = f(1)x^2.</math>
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We will prove by induction <math>f(x) = f(1)x^2</math> <math>\forall</math> <math>x.</math>
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If <math>x \le m</math> then <math>f(x) = f(1)x^2.</math> <math>\forall</math> <math>x</math> is true for some <math>m</math>.
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and if the statement is true for <math>m=k</math>
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<math>\Rightarrow f(k+1) =  f(k) + f(1) \pm 2\sqrt{f(k)f(1)} = f(1)(k+1)^2</math> or <math>f(1)(k-1)^2</math>
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and <math>f(k+1) =  f(k-1) + f(2) \pm 2\sqrt{f(k-1)f(2)} = f(1)(k+1)^2</math> or <math>f(1)(k-3)^2</math>
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<math>\Rightarrow f(k+1) = f(1)(k+1)^2.</math>
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<math>\Rightarrow</math> the statement is true for <math>m=k+1</math> as well.
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As the statement is true for <math>m = 4</math>, by mathematical induction we can conclude
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<math>f(x) = f(1)x^2</math> <math>\forall</math> <math>x.\Box</math>
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'''So, Case 2.1, Case 2.2.1 and Case 2.2.2 are the three independent possible solutions.'''
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 +
--Dineshram
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 +
==See Also==
 +
*[[IMO Problems and Solutions]]
 +
 
 +
{{IMO box|year=2012|num-b=3|num-a=5}}
 +
 
 +
[[Category:Olympiad Algebra Problems]]
 +
[[Category:Functional Equation Problems]]

Latest revision as of 07:08, 20 November 2023

Find all functions $f: \mathbb{Z} \to \mathbb{Z}$ such that, for all integers $a, b,$ and $c$ that satisfy $a + b+ c = 0$, the following equality holds: \[f(a)^2 + f(b)^2 + f(c)^2 = 2f(a)f(b) + 2f(b)f(c) + 2f(c)f(a).\] (Here $\mathbb{Z}$ denotes the set of integers.)

Solution

Consider $a = b = c = 0.$ Then $f(0)^2 + f(0)^2 + f(0)^2 = 2f(0)f(0) + 2f(0)f(0) + 2f(0)f(0) \Rightarrow 3f(0)^2 = 6f(0)^2 \Rightarrow$ \[f(0) = 0.\] Now we look at $b = -a, c = 0.$ $f(a)^2 + f(-a)^2 + f(0)^2 = 2f(a)f(-a) + 2f(-a)f(0) + 2f(0)f(a) \Rightarrow$ $f(a)^2 + f(-a)^2 = 2f(a)f(-a) \Rightarrow$ $f(a)^2 - 2f(a)f(-a) + f(-a)^2 = 0 \Rightarrow$ $(f(a) - f(-a))^2=0 \Rightarrow$ \[f(a)=f(-a).\]

We can write $f(c)^2 - 2f(c)(f(a)+f(b)) + (f(a)-f(b))^2 = 0 \Rightarrow$

\[f(c) = f(-c) = f(a+b) =\frac{2(f(a)+f(b)) \pm \sqrt{4(f(a)+f(b))^2 - 4(f(a)-f(b))^2}}{2}\]

\[\Rightarrow f(a+b) = f(a) + f(b) \pm 2\sqrt{f(a)f(b)}\]

If $f(b) = 0$, then \[f(a+b) = f(a) = f((a)mod(b))\]

Case 1: $f(1) = 0 \Rightarrow f(x)= 0$ $\forall$ $x.$ $\Box$

Case 2: $f(1) \not= 0$, we will have $f(2) = f(1) + f(1) \pm 2\sqrt{f(1)f(1)} \Rightarrow f(2) = 0$ or $f(2) = 4f(1)$

Case 2.1: $f(1) \not= 0, f(2) = 0 \Rightarrow f(x) = f(x$ $mod$ $2) \Rightarrow f(x) = f(1)$ if $x$ is odd, $f(x) = 0$ if $x$ is even. $\Box$

Case 2.2: $f(1) \not= 0, f(2) = 4f(1) \Rightarrow f(3) = f(2) + f(1) \pm 2\sqrt{f(2)f(1)}$

$\Rightarrow f(3) = 5f(1) \pm 4f(1) \Rightarrow f(3) = f(1)$ or $9f(1)$

Case 2.2.1: $f(1) \not= 0, f(2) = 4f(1), f(3) = f(1).$

$\Rightarrow f(4) = f(1) + f(3) \pm 2\sqrt{f(1)f(3)}$ and $f(4) = f(2) + f(2) \pm 2\sqrt{f(2)f(2)}$

$\Rightarrow f(4) = f(1)$ or $0$ and $f(4) = 16f(1)$ or $0$

$\Rightarrow f(4) = 0 \Rightarrow f(x) = f(x$ $mod$ $4).\Box$

Case 2.2.2: $f(1) \not= 0, f(2) = 4f(1), f(3) = 9f(1).$

$\Rightarrow f(4) = f(1 + 3) = f(1) + f(3) \pm 2\sqrt{f(1)f(3)} = 16f(1)$ or $4f(1)$

and $f(4) = f(2) + f(2) \pm 2\sqrt{f(2)f(2)} = 16f(1)$ or $0. \Rightarrow f(4) = 16f(1).$

If $x \le 4$ then $f(x) = f(1)x^2.$

We will prove by induction $f(x) = f(1)x^2$ $\forall$ $x.$

If $x \le m$ then $f(x) = f(1)x^2.$ $\forall$ $x$ is true for some $m$.

and if the statement is true for $m=k$

$\Rightarrow f(k+1) = f(k) + f(1) \pm 2\sqrt{f(k)f(1)} = f(1)(k+1)^2$ or $f(1)(k-1)^2$

and $f(k+1) = f(k-1) + f(2) \pm 2\sqrt{f(k-1)f(2)} = f(1)(k+1)^2$ or $f(1)(k-3)^2$

$\Rightarrow f(k+1) = f(1)(k+1)^2.$

$\Rightarrow$ the statement is true for $m=k+1$ as well.

As the statement is true for $m = 4$, by mathematical induction we can conclude

$f(x) = f(1)x^2$ $\forall$ $x.\Box$


So, Case 2.1, Case 2.2.1 and Case 2.2.2 are the three independent possible solutions.

--Dineshram

See Also

2012 IMO (Problems) • Resources
Preceded by
Problem 3 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 5
All IMO Problems and Solutions
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