Added geometric optimization
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Advanced/Geometric Optimization/main.tex
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251
Advanced/Geometric Optimization/main.tex
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% use [nosolutions] flag to hide solutions.
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% use [solutions] flag to show solutions.
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\documentclass[
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solutions
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]{../../resources/ormc_handout}
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\usepackage{../../resources/macros}
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\uptitlel{Advanced 2}
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\uptitler{\smallurl{}}
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\title{Geometric Optimization}
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\subtitle{
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Prepared by Mark on \today \par
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Based on a handout by Nakul \& Andreas
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}
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\begin{document}
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\maketitle
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\section{Optimization}
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\problem{}<simtri>
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Let $A$ and $B$ be two points on the same side of a given line $\ell$. \par
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Find a point $C$ on $\ell$ so that $|AC| + |BC|$ is minimized.
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\begin{center}
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\begin{tikzpicture}[scale = 2]
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\draw[-] (-2,0) -- (3,0);
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\fill[fill=black] (-0.6, 1) circle (0.03) node[below] {$A$};
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\fill[fill=black] (1.6, 0.75) circle (0.03) node[below] {$B$};
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\fill[fill=black] (0.5, 0) circle (0.03) node[below] {$C$};
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\draw[-] (-0.6, 1) -- (0.5, 0) -- (1.6, 0.75);
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\end{tikzpicture}
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\end{center}
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\vfill
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\pagebreak
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\definition{}
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An \textit{ellipse} with foci $A$, $B$ and radius $r$ is the set of all points $C$ where $|AB| + |BC| = r$.
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\problem{}
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Consider a reflective ellipse with foci $A$ and $B$. \par
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Find all points $X$ on the ellipse where $A$ can aim a laser at so that the beam reaches $B$. \par
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\hint{use \ref{simtri}}
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\begin{center}
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\begin{tikzpicture}[
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dot/.style={draw, fill, circle, inner sep=1.2},
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scale = 0.75
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]
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\def\a{5} % large half axis
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\def\b{3} % small half axis
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\draw (0,0) ellipse ({\a} and {\b});
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% Foci
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\node[dot,label={above right:$A$}] (A) at ({-sqrt(\a*\a-\b*\b)},0) {};
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\node[dot,label={above:$B$}] (B) at ({+sqrt(\a*\a-\b*\b)},0) {};
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% Node on ellipse
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\def\angle{150}
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\node[dot,label={\angle:$X$}] (X) at (\angle:{\a} and {\b}) {};
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\draw (A) -- (X) -- (B);
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\end{tikzpicture}
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\end{center}
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\vfill
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\pagebreak
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\problem{}
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Let $C$ be a point in the interior of a given angle. Find points A and B on the sides
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of the angle such that the perimeter of the triangle ABC is a minimum.
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\vfill
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\problem{}
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In a convex quadrilateral ABCD, find the point T for which the sum of the distances to
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the vertices is minimal.
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\vfill
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\pagebreak
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\problem{}
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A road needs to be constructed from town A to town B, crossing a river, over which a
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perpendicular bridge is to be constructed.
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Where should the bridge be placed to minimize $|AR_1| + |R_1R_2| + |R_2B|$?
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\begin{center}
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\begin{tikzpicture}[scale = 1.5]
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\draw[-] (-5, 0.5) -- (5, 0.5);
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\draw[-] (-5, -0.5) -- (5, -0.5);
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\fill[fill=black] (-3, -3) circle (0.06) node[below] {$A$};
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\fill[fill=black] (0.5, -0.5) circle (0.06) node[below] {$R_1$};
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\fill[fill=black] (0.5, 0.5) circle (0.06) node[below right] {$R_2$};
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\fill[fill=black] (3, 1) circle (0.06) node[below] {$B$};
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\draw[-] (-3, -3) -- (0.5, -0.5) -- (0.5, 0.5) -- (3,1);
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\end{tikzpicture}
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\end{center}
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\pagebreak
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\problem{}<equi>
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Consider an equilateral triangle triangle with vertices labeled $A$, $B$, and $C$. \par
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Let P be a point inside this triangle. Place $D$, $E$, and $F$ so that $PD$, $PE$, and $PF$
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are the perpendiculars from $P$ to the sides of the triangle. \par
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Find all points $P$ where $|PD| + |PE| + |PF|$ is minimized.
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\begin{center}
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\begin{tikzpicture}[scale = 4]
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\draw[-]
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(-1, 0)
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-- (1, 0)
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-- (0, 1.47)
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-- cycle
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;
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\fill[fill=black] (0, 1.47) circle (0.03) node[above] (A) {$A$};
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\fill[fill=black] (1, 0) circle (0.03) node[below right] {$B$};
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\fill[fill=black] (-1, 0) circle (0.03) node[below left] {$C$};
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\fill[fill=black] (0.39, 0.9) circle (0.03) node[above right] {$D$};
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\fill[fill=black] (-0.3, 0) circle (0.03) node[below right] {$E$};
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\fill[fill=black] (-0.555, 0.65) circle (0.03) node[above left] {$F$};
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\fill[fill=black] (-0.3, 0.5) circle (0.03) node[below right] {$P$};
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\draw[-] (-0.3, 0.5) -- (0.39, 0.9);
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\draw[-] (-0.3, 0.5) -- (-0.3, 0);
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\draw[-] (-0.3, 0.5) -- (-0.555, 0.65);
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\draw[-] (-0.2, 0) -- (-0.2, 0.1) -- (-0.3, 0.1);
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\end{tikzpicture}
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\end{center}
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\vfill
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\pagebreak
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\problem{}<equi2>
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With the same setup as \ref{equi}, find all points $P$ where $|PA| + |PB| + |PC|$ is minimized.
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\vfill
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\problem{}
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Solve \ref{equi2} for a triangle that isn't equilateral.
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\vfill
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\pagebreak
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\problem{}
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Draw a circle, then draw two distinct tangents $\ell_1$ and $\ell_2$ that intersect at point $A$. \par
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Let $P$ be a point on the circle between the tangents, and $BC$ be the tangent at that point.
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Describe how $P$ shoud be selected in order to minimize the perimeter of triangle $ABC$.
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\begin{center}
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\begin{tikzpicture}[scale = 2]
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\draw (0, 0) circle (1);
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\fill[fill=black] (-4, -1) circle (0.04) node[below] (A) {$A$};
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\draw[-] (-4, -1) -- (2, -1);
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\draw[-] (-4, -1) -- (1.2, 1.78);
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\draw[-] (-1, -1) -- (-1, 0.6);
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\fill[fill=black] (-1, 0.6) circle (0.04) node[above left] {$B$};
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\fill[fill=black] (-1, 0) circle (0.04) node[right] {$P$};
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\fill[fill=black] (-1, -1) circle (0.04) node[below] {$C$};
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\end{tikzpicture}
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\end{center}
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\vspace{2cm}
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\problem{}
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Now, assume that $\ell_1$ and $\ell_2$ intersect at $A$, and pick a point $P$ between them. \par
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Find $BC$ through $P$ so that the perimeter of $ABC$ is minimized.
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\vfill
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\pagebreak
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\section{Bonus Problems}
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\problem{}
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Given a cube $A_1B_1C_1D_1A_2B_2C_2D_2$ with side length $l$, \par
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find the angle and distance between lines $A_1B_2$ and $A_2C_1$.
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\begin{solution}
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Triangle $A_1B_2D_2$ is equilateral. \par
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Also, point $A_2$ is equidistant from each of this triangle's vertices. \par
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Therefore, its projection onto the plane formed by $A_1$, $B_2$, and $D_2$ is the center of the triangle. \par
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\vspace{2mm}
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Similarly, $C_1$ is mapped to the center of $A_1B_2D_2$. \par
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Therefore, lines $A_1B_2$ and $A_2C_1$ are perpendicular and the distance between them is
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equal to the distance from the center of triangle $A_1B_2D_2$ to its side.
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\vspace{2mm}
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Since all the sides of this triangle have length $l\sqrt{2}$, the distance in question is $\frac{a}{\sqrt{6}}$.
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\end{solution}
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\vfill
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\problem{}
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Consider a cube $A_1B_1C_1D_1A_2B_2C_2D_2$, and let $K$, $L$, \par
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and $M$ be midpoints of the edges $A_2D_2$, $A_1B_1$, and $C_1C_2$. \par
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Show that the triangle formed by $KLM$ is equilateral, and that its center is the center of the cube.
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\begin{solution}
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Let $O$ be the center of the cube. Then, $|OK| = |C_1D_2|$, $|2OL| = |D_2A_1|$, and $2|OM| = |A_1C_1|$. \par
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Since triangle $C_1D_2A_1$is equilateral, triangle $KLM$ is equilateral and has $O$ as its center.
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\end{solution}
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\vfill
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\pagebreak
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\problem{}
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Consider all $n$-gons with a certain perimeter.
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Show that the $n$-gon with maximal area has equal sides
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\vfill
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\problem{}
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Consider all $n$-gons with a certain perimeter.
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Show that the $n$-gon with maximal area has equal angles
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\vfill
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\end{document}
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