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Difference between revisions of "2022 SSMO Speed Round Problems/Problem 10"

(Created page with "==Problem== In a circle centered at <math>O</math> with radius <math>7,</math> we have non-intersecting chords <math>AB</math> and <math>DC.</math> <math>O</math> is outisde o...")
 
 
(One intermediate revision by the same user not shown)
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==Problem==
 
==Problem==
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In a circle centered at <math>O</math> with radius <math>7,</math> we have non-intersecting chords <math>AB</math> and <math>DC.</math> <math>O</math> is outisde of quadrilateral <math>ABCD</math> and <math>AB<CD.</math> Let <math>X = AO\cup CD</math> and <math>Y = BO\cup CD.</math> Suppose that <math>XO+YO = 7</math>. If <math>YC-DX=2</math> and <math>XY = 3</math>, then <math>AB = \frac{a\sqrt{b}}{c}</math> for <math>\gcd(a,c) = 1</math> and squareless <math>b.</math> Find <math>a+b+c.</math>
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Let <math>S = \{2,7,15,26,....\}</math> be the set of all numbers for which the <math>i^{th}</math> element in <math>S</math> is the sum of the <math>i^{th}</math> triangular number and the <math>i^{th}</math> positive perfect square. Let <math>T</math> be the set which contains all unique remainders when the elements in <math>S</math> are divided by <math>2022</math>. Find the number of elements in <math>T</math>.
  
 
==Solution==
 
==Solution==
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Let <math>CX = x+2, DY=x</math>.
 
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Then, by power of the point we have that
 
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<cmath>(x+2)(x+3) = 49 - YO^2 </cmath>
 
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<cmath>(x+5)x = 49 - XO^2</cmath>
 
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and subtracting gives that <math>XO^2 - YO^2 = 6</math>.
 
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Since we know that <math>XO + YO = 7</math>, dividing gives
 
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that <math>XO - YO = \frac{6}{7}</math> so <math>XO = \frac{55}{14}</math>
 
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and <math>YO = \frac{43}{14}</math>.
 
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Then, by law of cosines, it follows that
 
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<cmath>
 
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    OY^2 + OX^2 - 2 \cdot OX \cdot OY \cos\angle YOX = YX^2
 
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</cmath>
 
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which implies that <math>\cos\angle YOX = \frac{311}{473}</math>.
 
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Then, <math>AB^2 = OA^2 + OB^2 - 2 \cdot OA \cdot OB \cos\angle YOX</math>
 
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which implies that <math>AB = \frac{126 \sqrt{473}}{473}</math> so the
 
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answer is then <math>\boxed{1072}</math>.
 

Latest revision as of 19:14, 2 May 2025

Problem

Let $S = \{2,7,15,26,....\}$ be the set of all numbers for which the $i^{th}$ element in $S$ is the sum of the $i^{th}$ triangular number and the $i^{th}$ positive perfect square. Let $T$ be the set which contains all unique remainders when the elements in $S$ are divided by $2022$. Find the number of elements in $T$.

Solution

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