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2022 SSMO Speed Round Problems/Problem 4: Difference between revisions

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==Problem==
==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>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>
 
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==
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>


Suppose that <math>a_1 < a_2 < \dots < a_n</math> are taken such
<math>a_1 + a_2 + \dots + a_n</math> is minimal and
<cmath>
    F_{a_1} + F_{a_2} + \dots + F_{a_n} = 2023
</cmath>
Then, there are no consecutive <math>a_i</math> and <math>a_{i+1}</math> such
<math>a_i + 1 = a_{i+1}</math>.
Generalize <math>2023</math> to any <math>k > 1</math>.
Let <math>a</math> be maximal such <math>F_a < k</math>
We claim that <math>a_n = a</math>.
If <math>k</math> is a Fibonacci number the result follows.
Note that
<cmath>
    F_{2n+1} = F_{2n} + F_{2n-2} + \dots + F_2 + 1
</cmath>
and
<cmath>
    F_{2n+2} = F_{2n+1} + F_{2n-1} + \dots + F_1
</cmath>
follow inductively.
Thus, if <math>a_n \ne a</math> then
<cmath>
    F_{a_1} + F_{a_2} + \dots + F_{a_n} \le F_a < k,
</cmath>
contradiction.


Then, <math>a_n</math> be must maximal so we can reduce greedily to get
~pinkpig
<cmath>
    2023 = F_{17} + F_{14} + F_9 + F_7 + F_3
</cmath>
and the answer is <math>17 + 14 + 9 + 7 + 3 = \boxed{50}</math>

Latest revision as of 16:37, 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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