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...) |
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== Solution == | == Solution == | ||
− | {{ | + | Equality is equivalent to |
+ | <math> a^2 - ac + c^2 = b^2 + bd + d^2 (1) </math>. | ||
+ | |||
+ | 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 = | ||
+ | 60^\circ</math>, and <math> \angle BCD = 120^\circ</math>. Such a quadrilateral exists by <math>(1)</math> and the Law of Cosines. | ||
+ | |||
+ | By Strong Form of Ptolemy's Theorem, we find that; | ||
+ | |||
+ | <math>BD^2 = \frac{(ab+cd)(ad+bc)}{ac+bd}</math> | ||
+ | |||
+ | and by rearrangement inequality; | ||
+ | |||
+ | <math>ab+cd > ac+bd > ad+bc</math>. | ||
+ | |||
+ | 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 03:44, 25 October 2022
Problem
Let be positive integers and suppose that
Prove that is not prime.
Solution
Equality is equivalent to .
Let be the quadrilateral with , , , , , and . Such a quadrilateral exists by and the Law of Cosines.
By Strong Form of Ptolemy's Theorem, we find that;
and by rearrangement inequality;
.
Assume is a prime, since is an integer must be an integer but this is false since and . Thus can not be a prime.