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

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

@@ Line 1: / Line 1: @@
-Ummm. I'm only a middle school student and this is my first post, so i hope that i don't get this wrong ^_^
+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:
+<cmath>f(a)^2 + f(b)^2 + f(c)^2 = 2f(a)f(b) + 2f(b)f(c) + 2f(c)f(a).</cmath>
+(Here <math>\mathbb{Z}</math> denotes the set of integers.)
-Lets draw an circumcircle around triangle ABC (= circle ''a''), a circle with it's center as A and radius as AC (= circle ''b''),
+==Solution==
-a circle with it's center as B and radius as BC (= circle ''c'').
+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>
-Since the center of ''a'' lies on the line BC the three circles above are coaxial to line CD.
+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>
-Let ) Line AX and Line BX collide with ''a'' on P (not A) and Q (not B). Also let R be the point where AQ and BP intersects.
-Then since angle AYB = angle AZB = 90, by ceva's theorem in the opposite way, the point R lies on the line CD.
-Since triangles ABC and ACD are similar, AL^2 = AC^2 = AD X AB, so angle ALD = angle ABL
+We can write <math>f(c)^2 - 2f(c)(f(a)+f(b)) + (f(a)-f(b))^2 = 0  \Rightarrow</math>
-In the same way angle BKD = angle BAK
-So in total because angle ARD = angle ABQ = angle ALD, (A, R, L, D) is concyclic
-In the same way (B, R, K, D) is concyclic
-So angle ADR = angle ALR = 90, and in the same way angle BKR = 90 so the line RK and RL are tangent to each ''c'' and ''b''.
-Since R is on the line CD, and the line CD is the concentric line of ''b'' and ''c'', the equation RK^2 = RL^2 is true.
-Which makes the result of RK = RL. Since RM is in the middle and angle ADR = angle BKR = 90,
-we can say that the triangles RKM and RLM are the same. So KM = LM.
-Thanks for reading my first post!
+<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>
-by 장성광
-[[File:imo day 2 problem 4.jpg]]
+<cmath>\Rightarrow f(a+b) = f(a) + f(b) \pm 2\sqrt{f(a)f(b)}</cmath>
+If <math>f(b) = 0</math>, then <cmath>f(a+b) = f(a) = f((a)mod(b))</cmath>
+'''Case 1''': <math>f(1) = 0  \Rightarrow f(x)= 0</math> <math>\forall</math> <math>x.</math> <math>\Box</math>
+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>
+'''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>
+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>
+<math> \Rightarrow f(3) = 5f(1) \pm 4f(1) \Rightarrow f(3) = f(1)</math> or <math>9f(1)</math>
+'''Case 2.2.1''': <math>f(1) \not= 0, f(2) = 4f(1), f(3) = f(1).</math>
+<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>
+<math>\Rightarrow f(4) = f(1)</math> or <math>0</math> and <math>f(4) = 16f(1)</math> or <math>0</math>
+<math>\Rightarrow f(4) = 0 \Rightarrow  f(x) = f(x</math> <math>mod</math> <math>4).\Box</math>
+'''Case 2.2.2''': <math>f(1) \not= 0, f(2) = 4f(1), f(3) = 9f(1).</math>
+<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>
+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>
+If <math>x \le 4</math> then <math>f(x) = f(1)x^2.</math>
+We will prove by induction <math>f(x) = f(1)x^2</math> <math>\forall</math> <math>x.</math>
+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>.
+and if the statement is true for <math>m=k</math>
+<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>
+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>
+<math>\Rightarrow f(k+1) = f(1)(k+1)^2.</math>
+<math>\Rightarrow</math> the statement is true for <math>m=k+1</math> as well.
+As the statement is true for <math>m = 4</math>, by mathematical induction we can conclude
+<math>f(x) = f(1)x^2</math> <math>\forall</math> <math>x.\Box</math>
+'''So, Case 2.1, Case 2.2.1 and Case 2.2.2 are the three independent possible solutions.'''
+--Dineshram
+==See Also==
+*[[IMO Problems and Solutions]]
+{{IMO box|year=2012|num-b=3|num-a=5}}
+[[Category:Olympiad Algebra Problems]]
+[[Category:Functional Equation Problems]]

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

Page

Toolbox

Search

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

Latest revision as of 07:08, 20 November 2023

Solution

See Also