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Daily Insights from Magic Internet Math courses. Learn at https://mathacademy-cyan.vercel.app
๐Ÿ“– Definition II.2 (Gnomon) In any parallelogrammic area let any one whatever of the parallelograms about its diagonal with the two complements be called a gnomon. From: Euclid's Elements Learn more: Explore all courses:
๐Ÿ“ Fundamental Theorem of Arithmetic Every integer $n > 1$ can be written uniquely as a product of prime powers: $n = p_1^{a_1} p_2^{a_2} \\cdots p_k^{a_k}$ where $p_1 < p_2 < \\cdots < p_k$ are primes and $a_i \\geq 1$. From: df-course Learn more: Explore all courses:
๐Ÿ“– Coset For $H \\le G$ and $a \\in G$, the \\textbf{left coset} of $H$ containing $a$ is $aH = \\{ah : h \\in H\\}$. The \\textbf{right coset} is $Ha = \\{ha : h \\in H\\}$. From: df-course Learn more: Explore all courses:
๐Ÿ’ก Proposition III.32 (Tangent-Chord Angle) If a straight line touch a circle, and from the point of contact there be drawn across, in the circle, a straight line cutting the circle, the angles which it makes with the tangent will be equal to the angles in the alternate segments of the circle. From: Euclid's Elements Learn more: Explore all courses:
๐Ÿ“ Converse of Pythagorean Theorem (Euclid I.48) If in a triangle $a^2 + b^2 = c^2$, then the angle opposite the side of length $c$ is a right angle. Proof: Given $\\triangle ABC$ with $|AC|^2 + |BC|^2 = |AB|^2$. Construct right triangle $\\triangle DEF$ with legs $|DE| = |AC|$, $|EF| = |BC|$, right angle at $E$. By the Pythagorean theorem, $|DF|^2 = |DE|^2 + |EF|^2 = |AC|^2 + |BC|^2 = |AB|^2$. So $|DF| = |AB|$. By SSS congruence, $\\triangle ABC \\c... From: Four Pillars of Geometry Learn more: Explore all courses:
๐Ÿ“ Cyclic Quadrilateral Theorem A quadrilateral is inscribed in a circle if and only if its opposite angles sum to $180ยฐ$. Proof: Forward: Let $ABCD$ be inscribed in a circle. $\\angle A$ subtends arc $BCD$, $\\angle C$ subtends arc $DAB$. Together these arcs form the whole circle ($360ยฐ$ central angle). So $\\angle A + \\angle C = \\frac{1}{2}(360ยฐ) = 180ยฐ$. Similarly $\\angle B + \\angle D = 180ยฐ$. From: Four Pillars of Geometry Learn more: Explore all courses:
๐Ÿ“ Thales If parallel lines cut two transversals, then the segments cut off on one transversal are proportional to the corresponding segments on the other. That is, if $PQ \\parallel RS$ and they cut transversals at $A, P, B$ and $A, R, C$, then $\\frac{|AP|}{|PB|} = \\frac{|AR|}{|RC|}$. Proof: Consider triangles $\\triangle APR$ and $\\triangle BPS$. The parallel lines create equal corresponding angles. By similarity of triangles, the sides are proportional: $\\frac{|AP|}{|PB|} = \\frac{|AR|}{|RC|}$. From: Four Pillars of Geometry Learn more: Explore all courses:
๐Ÿ“ Matroid Greedy Algorithm A set system $(E, \\mathcal{I})$ is a matroid if and only if the greedy algorithm finds a maximum-weight independent set for all weight functions. Proof: If matroid: greedy builds an independent set $I$ by adding highest-weight elements that maintain independence. By the exchange property, $|I|$ equals the rank. Any independent set of same size has $\\leq$ weight (greedy chose highest weights). Conversely, if greedy works for all weights, the exch... From: Introduction to Graph Theory Learn more: Explore all courses:
๐Ÿ“– Euclid If a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the two lines, if extended indefinitely, meet on that side. From: Four Pillars of Geometry Learn more: Explore all courses:
๐Ÿ“– Family-Wise Error Rate (FWER) FWER = $\\Pr(\\text{at least one Type I error})$. When testing $m$ hypotheses at level $\\alpha$, FWER can be much larger than $\\alpha$. From: Intro to Statistical Learning Learn more: Explore all courses:
๐Ÿ“ Sample Theorem If $A \\subseteq B$ and $B \\subseteq A$, then $A = B$ Proof: Let $x \\in A$. Since $A \\subseteq B$, we have $x \\in B$ by definition of subset. Therefore, every element of $A$ is in $B$. Now, let $y \\in B$. Since $B \\subseteq A$, we have $y \\in A$ by definition. Therefore, every element of $B$ is in $A$. Since $A \\subseteq B$ and $B \\subseteq A... From: Algebraic Number Theory Learn more: Explore all courses:
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