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Difference between revisions of "2024 SSMO Team Round Problems/Problem 10"

(Created page with "==Problem== The side-lengths of a convex cyclic quadrilateral <math>ABCD</math> are integers and <math>(AB\cdot AD+BC\cdot CD)^2=AC^2\cdot BD^2-72</math>. Find all possible v...")
 
 
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==Solution==
 
==Solution==
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Let <math>AB = a,BC = b,CD = c,DA = d.</math> From Ptolemy's Theorem, we have <math>BC\cdot CD = ac+bd.</math> So, the equation is equivalent to
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\begin{align*}
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(ad+bc)^2 &= (ac+bd)^2-72\implies\\
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72&=a^2c^2+b^2d^2-a^2d^2-b^2c^2\implies\\
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72&=(a^2-b^2)(c^2-d^2).
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\end{align*}
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WLOG, assume that <math>a^2-b^2>c^2-d^2.</math> Since the difference of two perfect squares cannot be <math>2\pmod{4},</math> we have <math>(a^2-b^2,c^2-d^2) = (72,1),(24,3),(8,9).</math> Clearly, <math>(72,1)</math> has no solutions, as <math>c^2-d^2 = 1\implies c = 1,d = 0.</math> For <math>(24,3),</math> we have <math>c^2-d^2 = 3\implies c = 2,d = 1</math> and <math>a^2-b^2 = 24\implies (a-b)(a+b) = 24.</math>  This means <math>(a-b,a+b) = (2,12),(4,6)\implies (a,b) = (7,5),(5,1).</math> Finally, for <math>(8,9),</math> we have <math>c^2-d^2 = 9\implies c = 5,d = 4</math> and <math>a^2-b^2 = 8\implies a = 3,b = 1.</math> This gives <math>(a,b,c,d) = (7,5,2,1),(5,1,2,1),(3,1,5,4).</math> It is easy to verify that all three of these possibilities work, giving an answer of <cmath>(7+5+2+1)+(5+1+2+1)+(3+1+5+4) = 15+9+13 = \boxed{37}.</cmath>
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~SMO_Team

Latest revision as of 14:39, 10 September 2025

Problem

The side-lengths of a convex cyclic quadrilateral $ABCD$ are integers and $(AB\cdot AD+BC\cdot CD)^2=AC^2\cdot BD^2-72$. Find all possible values of the perimeter of $ABCD$.

Solution

Let $AB = a,BC = b,CD = c,DA = d.$ From Ptolemy's Theorem, we have $BC\cdot CD = ac+bd.$ So, the equation is equivalent to \begin{align*} (ad+bc)^2 &= (ac+bd)^2-72\implies\\ 72&=a^2c^2+b^2d^2-a^2d^2-b^2c^2\implies\\ 72&=(a^2-b^2)(c^2-d^2). \end{align*} WLOG, assume that $a^2-b^2>c^2-d^2.$ Since the difference of two perfect squares cannot be $2\pmod{4},$ we have $(a^2-b^2,c^2-d^2) = (72,1),(24,3),(8,9).$ Clearly, $(72,1)$ has no solutions, as $c^2-d^2 = 1\implies c = 1,d = 0.$ For $(24,3),$ we have $c^2-d^2 = 3\implies c = 2,d = 1$ and $a^2-b^2 = 24\implies (a-b)(a+b) = 24.$ This means $(a-b,a+b) = (2,12),(4,6)\implies (a,b) = (7,5),(5,1).$ Finally, for $(8,9),$ we have $c^2-d^2 = 9\implies c = 5,d = 4$ and $a^2-b^2 = 8\implies a = 3,b = 1.$ This gives $(a,b,c,d) = (7,5,2,1),(5,1,2,1),(3,1,5,4).$ It is easy to verify that all three of these possibilities work, giving an answer of \[(7+5+2+1)+(5+1+2+1)+(3+1+5+4) = 15+9+13 = \boxed{37}.\]

~SMO_Team

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