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Difference between revisions of "2001 IMO Shortlist Problems/N5"

(New page: == Problem == Let <math>a > b > c > d</math> be positive integers and suppose that <center><math>ac + bd = (b + d + a - c)(b + d - a + c).</math></center> Prove that <math>ab + cd</math> i...)
 
(→‎Solution)
 
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== Solution ==
 
== Solution ==
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{{solution}}
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Equality is equivalent to
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<math> a^2 - ac + c^2 = b^2 + bd + d^2  (1) </math>.
 +
 
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Let <math>ABCD</math> be the quadrilateral with <math>AB = a</math>, <math>BC = d</math>, <math>CD = b</math>, <math>AD = c</math>, <math> \angle BAD =
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60^\circ</math>, and <math> \angle BCD = 120^\circ</math>. Such a quadrilateral exists by <math>(1)</math> and the Law of Cosines.
 +
 
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By Strong Form of Ptolemy's Theorem, we find that;
 +
 
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<math>BD^2 = \frac{(ab+cd)(ad+bc)}{ac+bd}</math>
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and by rearrangement inequality;
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<math>ab+cd > ac+bd > ad+bc</math>.
 +
 
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Assume <math>ab+cd = p</math> is a prime, since <math>a^2 - ac + c^2 = BD^2</math> is an integer <math>p \times \frac{ad+bc}{ac+bd}</math> must be an integer but this is false since <math>(p,ac+bd) = 1</math> and <math>ac+bd > ad+bc</math>. Thus <math>ab+cd</math> can not be a prime.
  
 
== Resources ==
 
== Resources ==

Latest revision as of 04:44, 25 October 2022

Problem

Let $a > b > c > d$ be positive integers and suppose that

$ac + bd = (b + d + a - c)(b + d - a + c).$

Prove that $ab + cd$ is not prime.

Solution

Equality is equivalent to $a^2 - ac + c^2 = b^2 + bd + d^2 (1)$.

Let $ABCD$ be the quadrilateral with $AB = a$, $BC = d$, $CD = b$, $AD = c$, $\angle BAD = 60^\circ$, and $\angle BCD = 120^\circ$. Such a quadrilateral exists by $(1)$ and the Law of Cosines.

By Strong Form of Ptolemy's Theorem, we find that;

$BD^2 = \frac{(ab+cd)(ad+bc)}{ac+bd}$

and by rearrangement inequality;

$ab+cd > ac+bd > ad+bc$.

Assume $ab+cd = p$ is a prime, since $a^2 - ac + c^2 = BD^2$ is an integer $p \times \frac{ad+bc}{ac+bd}$ must be an integer but this is false since $(p,ac+bd) = 1$ and $ac+bd > ad+bc$. Thus $ab+cd$ can not be a prime.

Resources

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