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Difference between revisions of "2023 SSMO Speed Round Problems/Problem 5"

(Created page with "==Problem== Let <math>F_1 = F_2 = 1</math> and <math>F_n = F_{n-1} + F_{n-2}</math> for all <math>n\geq 2</math> be the Fibonacci numbers. If distinct positive integers <math>...")
 
 
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==Problem==
 
==Problem==
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Let <math>F_1 = F_2 = 1</math> and <math>F_n = F_{n-1} + F_{n-2}</math> for all <math>n\geq 2</math> be the Fibonacci numbers. If distinct positive integers <math>a_1, a_2, \dots a_n</math> satisfies <math>F_{a_1}+F_{a_2}+\dots+F_{a_n}=2023</math>, find the minimum possible value of <math>a_1+a_2+\dots+a_n.</math>
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In a parallelogram <math>ABCD</math> of dimensions <math>6\times 8,</math> a point <math>P</math> is choosen such that <math>\angle{APD}+\angle{BPC} = 180^{\circ}.</math> Find the sum of the maximum, <math>M</math>, and minimum values of <math>(PA)(PC)+(PB)(PD).</math> If you think there is no maximum, let <math>M=0.</math>
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==Solution==
 
==Solution==
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A translation that takes <math>BC</math> to <math>AD</math> takes <math>P</math> to <math>P'.</math> Thus, <math>\angle{AP'D}+\angle{APD} = \angle{BPC}+\angle{APD} = 180^{\circ},</math> meaning <math>PAP'D</math> is cyclic. From Ptolemy's Theorem, <math>(AD)(PP') = (PA)(P'D)+(PD)(P'A) \implies (PA)(PC)+(PB)(PD)=(AD)(CD)=6\cdot8 = 48,</math> meaning the answer is <math>48+48 = \boxed{96}</math>
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Suppose that <math>a_1 < a_2 < \dots < a_n</math> are taken such
 
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<math>a_1 + a_2 + \dots + a_n</math> is minimal and
 
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<cmath>
 
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    F_{a_1} + F_{a_2} + \dots + F_{a_n} = 2023
 
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</cmath>
 
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Then, there are no consecutive <math>a_i</math> and <math>a_{i+1}</math> such
 
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<math>a_i + 1 = a_{i+1}</math>.
 
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Generalize <math>2023</math> to any <math>k > 1</math>.
 
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Let <math>a</math> be maximal such <math>F_a < k</math>
 
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We claim that <math>a_n = a</math>.
 
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If <math>k</math> is a Fibonacci number the result follows.
 
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Note that
 
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<cmath>
 
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    F_{2n+1} = F_{2n} + F_{2n-2} + \dots + F_2 + 1
 
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</cmath>
 
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and
 
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<cmath>
 
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    F_{2n+2} = F_{2n+1} + F_{2n-1} + \dots + F_1
 
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</cmath>
 
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follow inductively.
 
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Thus, if <math>a_n \ne a</math> then
 
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<cmath>
 
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    F_{a_1} + F_{a_2} + \dots + F_{a_n} \le F_a < k,
 
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</cmath>
 
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contradiction.
 
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Then, <math>a_n</math> be must maximal so we can reduce greedily to get
 
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<cmath>
 
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    2023 = F_{17} + F_{14} + F_9 + F_7 + F_3
 
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</cmath>
 
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and the answer is <math>17 + 14 + 9 + 7 + 3 = \boxed{50}</math>
 

Latest revision as of 14:20, 3 July 2023

Problem

In a parallelogram $ABCD$ of dimensions $6\times 8,$ a point $P$ is choosen such that $\angle{APD}+\angle{BPC} = 180^{\circ}.$ Find the sum of the maximum, $M$, and minimum values of $(PA)(PC)+(PB)(PD).$ If you think there is no maximum, let $M=0.$


Solution

A translation that takes $BC$ to $AD$ takes $P$ to $P'.$ Thus, $\angle{AP'D}+\angle{APD} = \angle{BPC}+\angle{APD} = 180^{\circ},$ meaning $PAP'D$ is cyclic. From Ptolemy's Theorem, $(AD)(PP') = (PA)(P'D)+(PD)(P'A) \implies (PA)(PC)+(PB)(PD)=(AD)(CD)=6\cdot8 = 48,$ meaning the answer is $48+48 = \boxed{96}$

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