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Difference between revisions of "2022 SSMO Speed Round Problems/Problem 4"

(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>...")
 
 
(2 intermediate revisions by the same user not shown)
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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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Consider a quadrilateral <math>ABCD</math> with area <math>120</math> and satisfying <math>AB+CD=AD+BC=24</math>. There exists a point <math>P</math> in 3D space such that the distances from <math>P</math> to <math>AB</math>, <math>BC</math>, <math>CD</math>, and <math>DA</math> are all equal to <math>13</math>. Find the volume of <math>PABCD</math>.
  
 
==Solution==
 
==Solution==
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Let <math>r</math> be the inradius of <math>ABCD</math>. Note that <cmath>\frac{1}{2}r(AB+BC+CD+DA)=[ABCD]\implies r = \frac{2\cdot120}{24+24} = 5.</cmath> We have <cmath>h^2+5^2 = 13^2\implies h = 12,</cmath> meaning the volume of <math>PABCD</math> is <cmath>\frac{1}{3}h[ABCD] = \frac{1}{3}(12)(120) = \boxed{480}.</cmath>
  
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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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~pinkpig
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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 17:37, 13 September 2025

Problem

Consider a quadrilateral $ABCD$ with area $120$ and satisfying $AB+CD=AD+BC=24$. There exists a point $P$ in 3D space such that the distances from $P$ to $AB$, $BC$, $CD$, and $DA$ are all equal to $13$. Find the volume of $PABCD$.

Solution

Let $r$ be the inradius of $ABCD$. Note that \[\frac{1}{2}r(AB+BC+CD+DA)=[ABCD]\implies r = \frac{2\cdot120}{24+24} = 5.\] We have \[h^2+5^2 = 13^2\implies h = 12,\] meaning the volume of $PABCD$ is \[\frac{1}{3}h[ABCD] = \frac{1}{3}(12)(120) = \boxed{480}.\]


~pinkpig

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