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Difference between revisions of "User:Quantum-phantom"
 
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This message is regarding [url=https://artofproblemsolving.com/community/c594864h3265596p30137934]this post[/url].
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By the law of cosines,
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[quote="fura3334"][b]Problem 10[/b]
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<cmath>\cos\angle DAC=\frac{221-CD^2}{220}=\cos\angle DBC=\frac{169-CD^2}{120},</cmath>
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(unclear source)
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so <math>CD=\sqrt{\tfrac{533}{5}}</math>. Similarly, <math>AB=\tfrac{2 \sqrt{5} \sqrt{533}}{13}</math>. Let <math>AD\cap BC=I</math>, <math>AB\cap CD=J</math>, <math>OE\cap IJ=F</math>, then <math>OF\perp IJ</math> by Brocard's theorem. Since <math>ON\perp DC</math>, <math>OM\perp AB</math>, then
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For a positive integer <math>n</math>, denote by <math>f(n)</math> the number of ways to represent <math>n</math> as the sum of powers of 2. Different orders [color=#960000]are[/color] considered different representations, e.g. <math>4+1+1</math> and <math>1+4+1</math> are two different representations of <math>6</math>. Call <math>n</math> "good" if <math>f(n)</math> is even. Let <math>A</math> be the largest number of consecutive good numbers in <math>\{1,2,\cdots,2025\}</math>. Find the remainder when <math>A</math> is divided by 1000.[/quote]
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\begin{align*}
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\frac{MP}{NP}&=\frac{MO\sin\angle MOE}{NO\sin\angle NOE}=\frac{\sin\angle MOE}{\sin\angle NOE}=\frac{\sin\angle IJA}{\sin(\pi-\angle DJI)}=\frac{\sin\angle IJA}{\sin\angle DJI}\\
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I just read [color=#960000]are[/color] as [color=#960000]aren't[/color]. (yesterday)
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&=\frac{JA\cdot JI\sin\angle IJA}{IJ\cdot DJ\sin\angle DJI}\cdot\frac{DJ}{JA}=\frac{[IJA]}{[DJI]}\cdot\frac{DJ}{JA}=\frac{IA}{ID}\cdot\frac{DJ}{JA}.
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\end{align*}
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Let \(n=a_1+a_2+\cdots\) where all \(a_i\) are powers of \(2\). If \(a_1=1\), then \(a_2+a_3+\dots=n-1\), with \(f(n-1)\) choices; If \(a_1=2\), then \(a_2+a_3+\dots=n-2\), with \(f(n-2)\) choices... So \(f(n)=\sum\limits_{i=0}^{\left\lfloor\log_2n\right\rfloor}f\left(n-2^i\right)\).
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By the law of sines,
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<cmath>\frac{IA}{ID}=\frac{IA}{IC}\cdot\frac{IC}{ID}=\frac{AB}{CD}\cdot\frac{AC}{BD}=\frac{55}{78},~\frac{DJ}{JA}=\frac{BD}{AC}=\frac{12}{11}.</cmath>
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We use induction on \(n\) to show that <math>f(n)</math> is odd iff <math>n</math> is in the form of \(2^t-1\), where \(t\) is a natural number.
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So the answer is \(\tfrac{MP}{NP}=\frac{12}{11}\cdot\frac{55}{78}=\frac{10}{13}\).
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[img]https://img.picgo.net/2024/04/07/IMG_3135a148c4fcc55ace27.jpeg[/img]

Latest revision as of 05:11, 15 April 2024

By the law of cosines, \[\cos\angle DAC=\frac{221-CD^2}{220}=\cos\angle DBC=\frac{169-CD^2}{120},\] so $CD=\sqrt{\tfrac{533}{5}}$. Similarly, $AB=\tfrac{2 \sqrt{5} \sqrt{533}}{13}$. Let $AD\cap BC=I$, $AB\cap CD=J$, $OE\cap IJ=F$, then $OF\perp IJ$ by Brocard's theorem. Since $ON\perp DC$, $OM\perp AB$, then \begin{align*} \frac{MP}{NP}&=\frac{MO\sin\angle MOE}{NO\sin\angle NOE}=\frac{\sin\angle MOE}{\sin\angle NOE}=\frac{\sin\angle IJA}{\sin(\pi-\angle DJI)}=\frac{\sin\angle IJA}{\sin\angle DJI}\\ &=\frac{JA\cdot JI\sin\angle IJA}{IJ\cdot DJ\sin\angle DJI}\cdot\frac{DJ}{JA}=\frac{[IJA]}{[DJI]}\cdot\frac{DJ}{JA}=\frac{IA}{ID}\cdot\frac{DJ}{JA}. \end{align*} By the law of sines, \[\frac{IA}{ID}=\frac{IA}{IC}\cdot\frac{IC}{ID}=\frac{AB}{CD}\cdot\frac{AC}{BD}=\frac{55}{78},~\frac{DJ}{JA}=\frac{BD}{AC}=\frac{12}{11}.\] So the answer is \(\tfrac{MP}{NP}=\frac{12}{11}\cdot\frac{55}{78}=\frac{10}{13}\). [img]https://img.picgo.net/2024/04/07/IMG_3135a148c4fcc55ace27.jpeg[/img]

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