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Difference between revisions of "2012 IMO Problems/Problem 4"
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Ummm. I'm only a middle school student and this is my first post, so i hope that i don't get this wrong ^_^
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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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<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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(Here <math>\mathbb{Z}</math> denotes the set of integers.)
  
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Lets draw an circumcircle around triangle ABC (= circle ''a''), a circle with it's center as A and radius as AC (= circle ''b''),
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'''Solution:'''  
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a circle with it's center as B and radius as BC (= circle ''c'').
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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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Since the center of ''a'' lies on the line BC the three circles above are coaxial to line CD.
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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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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.
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What about <math>b = a, c = -2a?</math> Then <math>f(a)^2 + f(a)^2 + f(-2a)^2 = 2f(a)f(a) + 2f(a)f(-2a) + 2f(-2a)f(a) \Rightarrow</math> <math>2f(a)^2 + f(-2a)^2 = 2f(a)^2 + 4f(a)f(-2a) \Rightarrow</math> <math>f(-2a)^2 = 4f(a)f(-2a) \Rightarrow</math> <cmath>f(-2a) = f(2a) = 4f(a).</cmath>
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Then since angle AYB = angle AZB = 90, by ceva's theorem in the opposite way, the point R lies on the line CD.
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We conjecture that <math>f(na) = n^2f(a).</math> Consider <math>b = n\cdot{a}, c = -(n+1)\cdot{a}</math> and assume that <math>f(a) \neq 0.</math> If it does, we get that the constant 0 function satisfies the conditions of the problem.
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<math>f(a)^2 + f(n\cdot{a})^2 + f((n+1)\cdot{a})^2 =</math> <math>2f(a)f(n\cdot{a}) + 2f(n\cdot{a})f((n+1)\cdot{a}) + 2f((n+1)\cdot{a})f(a) \Rightarrow</math>
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Since triangles ABC and ACD are similar, AL^2 = AC^2 = AD X AB, so angle ALD = angle ABL
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<math>f(a)^2 + (n^2f(a))^2 + ((n+1)^2f(a))^2 =</math> <math>2n^2f(a)^2 + 2n^2(n+1)^2f(a)^2 + 2(n+1)^2f(a)^2 \Rightarrow</math>
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In the same way angle BKD = angle BAK
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<math>1 + n^4 + (n+1)^4 = 2n^2 + 2n^2(n+1)^2 +  2(n+1)^2 \Rightarrow</math>
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So in total because angle ARD = angle ABQ = angle ALD, (A, R, L, D) is concyclic
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<math>1 + n^4 + n^4 + 4n^3 + 6n^2 + 4n + 1 = 2n^2 + 2n^4 + 4n^3 + 2n^2 + 2n^2 + 4n +2 \Rightarrow</math>
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In the same way (B, R, K, D) is concyclic
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<math>2n^4 + 4n^3 + 6n^2 + 4n +2 = 2n^4 + 4n^3 + 6n^2 + 4n + 2 \hspace{7 mm} \checkmark</math>
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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''.
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<cmath>f(n\cdot{a}) = n^2f(a)</cmath>
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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.
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We note that <math>f(a)=a^2f(1).</math> This means that we want to find what the possible values of <math>f(1)</math> are in order to finish the problem. <math>a + b + c = 0 \Rightarrow</math> <math>c = -(a+b) \Rightarrow</math> <math>a^4f(1)^2 + b^4f(1)^2 + (a+b)^4f(1)^2 =</math> <math>2a^2b^2f(1)^2 + 2b^2(a+b)^2f(1)^2 + 2(a+b)^2a^2f(1)^2.</math> Returning to the above manipulations (the ones used to show that <math>f(n\cdot{a}) = n^2f(a)</math>), we see that letting <math>n = \frac{a}{b}</math> and multiplying through by <math>b^4</math> yields precisely this result (again, assuming that <math>f(1) \neq 0,</math> with equality yielding the fact that the constant 0 function satisfies the condition). Therefore, <math>f(1)</math> can be any integer value (since <math>f</math> maps the integers to the integers only), and setting <math>f(1) = m</math>, we see that any function of the form <cmath>f(a) = m\cdot{a^2}, m \in \mathbb{Z}</cmath> satisfies the condition.
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Which makes the result of RK = RL. Since RM is in the middle and angle ADR = angle BKR = 90,
 
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we can say that the triangles RKM and RLM are the same. So KM = LM.
 
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Thanks for reading my first post!
 
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by 장성광
 
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[[File:picture]]
 

Revision as of 20:02, 14 June 2013

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 $b = -a, c = 0.$ $f(a)^2 + f(-a)^2 + f(0)^2 = 2f(a)f(-a) + 2f(-a)f(0) + 2f(0)f(a) \Rightarrow$ $f(a)^2 + f(-a)^2 = 2f(a)f(-a) \Rightarrow$ $f(a)^2 - 2f(a)f(-a) + f(-a)^2 = 0 \Rightarrow$ $(f(a) - f(-a))^2=0 \Rightarrow$ \[f(a)=f(-a).\] What about $b = a, c = -2a?$ Then $f(a)^2 + f(a)^2 + f(-2a)^2 = 2f(a)f(a) + 2f(a)f(-2a) + 2f(-2a)f(a) \Rightarrow$ $2f(a)^2 + f(-2a)^2 = 2f(a)^2 + 4f(a)f(-2a) \Rightarrow$ $f(-2a)^2 = 4f(a)f(-2a) \Rightarrow$ \[f(-2a) = f(2a) = 4f(a).\] We conjecture that $f(na) = n^2f(a).$ Consider $b = n\cdot{a}, c = -(n+1)\cdot{a}$ and assume that $f(a) \neq 0.$ If it does, we get that the constant 0 function satisfies the conditions of the problem. $f(a)^2 + f(n\cdot{a})^2 + f((n+1)\cdot{a})^2 =$ $2f(a)f(n\cdot{a}) + 2f(n\cdot{a})f((n+1)\cdot{a}) + 2f((n+1)\cdot{a})f(a) \Rightarrow$ $f(a)^2 + (n^2f(a))^2 + ((n+1)^2f(a))^2 =$ $2n^2f(a)^2 + 2n^2(n+1)^2f(a)^2 + 2(n+1)^2f(a)^2 \Rightarrow$ $1 + n^4 + (n+1)^4 = 2n^2 + 2n^2(n+1)^2 + 2(n+1)^2 \Rightarrow$ $1 + n^4 + n^4 + 4n^3 + 6n^2 + 4n + 1 = 2n^2 + 2n^4 + 4n^3 + 2n^2 + 2n^2 + 4n +2 \Rightarrow$ $2n^4 + 4n^3 + 6n^2 + 4n +2 = 2n^4 + 4n^3 + 6n^2 + 4n + 2 \hspace{7 mm} \checkmark$ \[f(n\cdot{a}) = n^2f(a)\] We note that $f(a)=a^2f(1).$ This means that we want to find what the possible values of $f(1)$ are in order to finish the problem. $a + b + c = 0 \Rightarrow$ $c = -(a+b) \Rightarrow$ $a^4f(1)^2 + b^4f(1)^2 + (a+b)^4f(1)^2 =$ $2a^2b^2f(1)^2 + 2b^2(a+b)^2f(1)^2 + 2(a+b)^2a^2f(1)^2.$ Returning to the above manipulations (the ones used to show that $f(n\cdot{a}) = n^2f(a)$), we see that letting $n = \frac{a}{b}$ and multiplying through by $b^4$ yields precisely this result (again, assuming that $f(1) \neq 0,$ with equality yielding the fact that the constant 0 function satisfies the condition). Therefore, $f(1)$ can be any integer value (since $f$ maps the integers to the integers only), and setting $f(1) = m$, we see that any function of the form \[f(a) = m\cdot{a^2}, m \in \mathbb{Z}\] satisfies the condition.

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