Assignment 5
Problems
The Parallel Postulate
Recall that Playfair’s Axioms (already suggested by Proclus in the 400s) was a simpler re-phrasing of Euclid’s original fifth postulate on parallel (that is, nonintersecting) lines.
Proposition 1 Given any line \(L\) in \(\EE^2\), and any point \(p\in \EE^2\) not lying on \(L\), there exists a unique line \(\Lambda\) through \(p\) which does not intersect \(p\).
Exercise 1 Prove Proposition prp-proving-playfairs-axiom.
Hint: use isometries to help you out!
First, use an isometry to move \(L\) to the \(x\)-axis. Then, use another isometry to keep \(L\) on the \(x\) axis, but to move \(p\) to some point along the \(y\) axis. Then, prove that through any point on the \(y\) axis there is a unique line that does not intersect the \(x\)-axis.
Similarities and Lines
We saw in Theorem Isometries Send Lines to Lines that any isometry will carry a line to another line. The same is true more generally of similarities:
Exercise 2 (Similarities Send Lines to Lines) Let \(\gamma\colon\RR\to\EE^2\) be a line, and \(\sigma\colon\EE^2\to\EE^2\) be a similarity. Prove that \(\sigma\circ\gamma\) is also a line.
*Hint-replicate the proof of Theorem Isometries Send Lines to Lines as closely as possible, replacing the isometry \(\phi\) with the similarity \(\sigma\), and keeping track of the scaling factors of \(\sigma\) versus \(\sigma^{-1}\) (Proposition prp-similarity-scaling-inverse).
Distance to a Line.
In this problem it’s probably helpful to use the ‘calculus trick’ offered as an optional problem last week: that is, if you are looking to minimize a positive function \(f(x)\), you can instead try to minimize the function \(f(x)^2\), and you’ll find the same \(x\)-value achieves the minimum.
The reason this is useful to us is that the distance function in geometry has a square root in it, and differentiating roots can be a lot of work. So this says instead we can minimize the square of distance to find the right point.
Exercise 3 (Closest Point on Line) Let \(L\) be the line traced by the affine curve \(\gamma(t)=\pmat{a t+ c\\ bt +d}\), and \(O\) be the origin.
Alternatively, do this for the line \((2t+1,3t-4)\), to save yourself a lot of \(abcd's\).
- Find the point \(p\in L\) which is closest to \(O\).
- Calculate \(\dist(O,L)\).
- What angle does the segment connecting \(p\) to \(O\) make with the line \(L\)? (We haven’t reviewed the definition of angle yet, so just use your knowledge of angles from precalculus: pick a line an compute an example!)
This problem shows how we can use calculus as a tool to discover a geometric fact: here we learned something about where the closest point on a line is located (by finding it with calculus, then calculating the angle formed).
Often calculus can provide the tools needed for discovery of a new fact, but once its known, one can often go back and find a geometric (more in the style of Euclid) proof that all of a sudden makes the conclusion feel inevitable.
Exercise 4 (Closest Point: Geometric Reasoning) Now that we know the answer, formulate in your own words a geometric proposition that describes which point \(p\) on a line is closer to a given point \(q\) not on that line.
Prove this propostion without just taking a derivative, try to reason more like the Greeks, using other facts and theorems that we’ve proven.
Hint: if you pick some other point \(r\) along \(L\), can you draw a triangle using \(p,q,r\)? How can we use the geometry of this triangle to show that \(r\) is farther from \(q\) than \(p\) was? Does the Pythagorean theorem say anything useful?
Intersecting Circles
Recall that in all of Euclid’s axioms, conditions for intersections with circles were never specified! Indeed - Euclid intersected two circles in his construction of the equilateral triangle. Now that we have a precise description of circles in our new foundations, we can fix this gap:
Exercise 5 Prove that two circles intersect each other if the distance between their centers is less than or equal to the sum of their radii.
Alternatively: do this for the specific case where the circles radii are 3 and 4, and the distance between thier centers is 5.
Hint: start by applying an isometry to move one of the circles to have center \((0,0)\), and then another isometry to roate everything so the second circle has center \((x,0)\) along the \(x\)-axis. This will make computations easier!
Parabolas
A parabola was one of few curves that the Greeks understood. However, while we often think of a parabola algebraically that was not their original definition!
Definition 1 (Parabola) Given a point \(f\) (called the focus) and a line \(L\) not containing \(f\) (called the directrix), the parabola with focus \(f\) and directrix \(L\) is the curve \(C\) of points where for each \(p\in C\) the distance to the focus equals the distance to the line: \[\dist(p,f)=\dist(p,L)\]
Exercise 6 In this problem we confirm that \(y=x^2\) is indeed a parabola! Let \(L\) be a horizontal line intersecting the \(y-\)axis at some point \((0,-\ell)\), and \(f=(0,h)\) be a point along the \(y\)-axis for \(\ell,h>0\).
- Write down an algebraic equation for the coorddinates of a point \((x,y)\) determining when it is on the parabola with focus \(f\) and directrix \(L\).
- Find which point \(f\) and line \(L\) make this parabola have the algebraic equation \(y=x^2\).
Solutions
The Parallel Postulate
Let \(L\) be any line in the plane, and \(p\) a point not on that line. Choose some point \(q\) on the line \(L\). Then, by a translation isometry we know we can move \(q\) to the origin, so now our line passes through \(O\). Let \(v\) be the tangent vector to our line at \(O\). Now we may use a rotation isometry fixing \(O\) which takes \(v\) to \(\langle 1,0\rangle_O\). We know the \(x\) axis is a line through \(O\) whose tangent vector is \(\langle 1,0\rangle\), and we know that lines through any point/direction are unique: thus our line must be the \(x\)-axis!
This colleciton of isometries has moved our point \(p\) to some point not on the \(x\) axis (since it is not on the line!). Say \(p\) has moved to the point with coordinates \((h,k)\). Now, we may use a translation \((x,y)\mapsto (x-h,y)\) to translate this point onto the \(y\) axis. This horizontal translation preserves the \(x\) axis, so after this whole sequence of isometries we now have our line on the \(x\) axis and our point on the \(y\) axis at height \(h\).
It remains only to show that there is a unique line through \((0,h)\) which does not intersect the \(x\) axis. All lines are affine functions, so we may write an arbitrary line through \((0,h)\) as
\[\ell(t)=\pmat{v_1\\v_2}t+\pmat{0\\ h}\]
Every point on the \(x\) axis has its \(y\) coordinates equal to zero, so \(\ell\) intersects the \(x\) axis whenever \(v_2 t+h=0\). Solving this we get
\[t= -h/v_2\]
Which requires \(v_2\neq 0\): thus every line with nonzero \(v_2\) intersects the \(x-\)axis! Whenever \(v_2=0\) the direction vector of the line is horizontal, and so these all describe the same line - the horizontal line through \((0,h)\). Because this is the only case where we don’t find an intersection with the \(x\) axis, there is a unique line which doesn’t intersect the \(x\) axis, just as Proclus and Playfair claimed!
Similarities and Lines
We wish to show a similarity \(\sigma\) maps a line \(L\) to another line, following the same strategy as in the proof for isometries. Let \(k\) be the scaling factor of the similarity, and consider the curve \(\sigma\circ L(t)\). We wish to show that \(L\) is a line, so we want to show its a minimizing curve.
So, choose two points \(p,q\) along \(\sigma(L)\) and consider the segment between them. If this were not the shortest segment there’d be some shorter curve \(\alpha\), so we’d have \[\length(\alpha)<\length(\sigma(L))\]
But we know what \(\sigma\) does to lengths of curves: its scaling factor is \(k\), so it scales the length of everythign by \(k\). That is \[\length(\sigma(L))=k\length(L)\] Where by the length of \(L\) we mean the length of the segment of \(L\) which \(\sigma\) maps to the segment original from \(p\) to \(q\). Putting these two inequalities together we see \[\length(\alpha)<k\length(L)\implies \frac{1}{k}\length(\alpha)<\length(L)\]
But, since \(\sigma\) is a similarity with scaling factor \(k\), its inverse \(\sigma^{-1}\) is a similarity with scaling factor \(1/k\), and so \(\tfrac{1}{k}\length(\alpha)\) is the length of the curve \(\sigma^{-1}(\alpha)\). But now we’ve reached a contradiction! Because \(L\) is a line, so its supposed to be the shortest curve between any two points, and we just found a curve, \(\sigma^{-1}\circ\alpha\), that’s shorter!
Thus this whole situation must be impossible, and there is in fact no curve \(\alpha\) shorter than \(\sigma\circ L\). This makes \(\sigma\circ L\) the shortest, so its a line! Which is what we wanted to show in the first place.
Distance To A Line I
We are given a line as an affine equation \(\gamma(t)=(at+c, bt+d)\) and we wish to minimize the distance to the origin. We know the distance formula so we can compute the distance from \(\gamma(t)\) to \(O\) for any time:
\[\mathrm{dist}(\gamma(t),O)=\sqrt{(at+c)^2+(bt+d)^2}\]
So, we want to find the value of \(t\) that minimizes this function. The standard calculus trick would be to take the derivative and set it equal to zero: and this works just fine! Howver, if you did the optional problem last week, you remember that we can also minmize its square, which makes the algebra easier since that gets rid of the square root:
\[\mathrm{dist}^2=(at+c)^2+(bt+d)^2\]
Minimizing this also requires differentiating and setting equal to zero. This results in a lot of \(a,b,c\) and \(d\)’s (thanks, chain rule!) so for readability we’ll switch to the more specific version of the question, and look at the line \(\gamma(t)=(2t+1,3t-4)\). This means we are minimizing \(\mathrm{dist}^2=(2t+1)^2+(3t-4)^2\), which when differentiated yields
\[ \frac{d}{dt}\left((2t+1)^2+(3t+4)^2\right)=2\cdot 2(2t+1)+2\cdot 3(3t-4)=0 \]
This is a linear equation in \(t\), so we can just collect like terms and sovle!
\[8t+4+18t-24=0 \implies 26t=20\implies t=10/13\]
This tells us the time that we are at the location of the minimum, but not yet what the point, or what the minimum value are! We need to plug back into \(\gamma\) to recover the point, and then plug this into the distance formula:
\[\gamma(t_{\min})=\left(2\frac{10}{13}+1,3\frac{10}{13}-4\right)=\left(\frac{33}{13},\frac{-22}{13}\right)\]
\[\mathrm{dist}\left(\gamma(t_{\min}),O\right)=\sqrt{\left(\frac{33}{13}\right)^2+\left(\frac{22}{13}\right)^2}=\sqrt{\frac{1573}{13^2}}=\frac{11}{\sqrt{13}}\]
Now that we have the distance, we just want to understand the angle formed by connecting this point to the origin, and the original line. The slope of the original line is given by its derivative, and since \(\gamma(t)=(2t+1,3t-4)\) this means its direction is \(\gamma^\prime = (2,3)\).
To find the slope of the line connecting our new point to the origin, we can simplify things a bit by multiplying our minimizing point by various scalars (multiplying by a constant wont change a slope): so we multiply by 13 to get \((33,-22)\) and then we divide by 11 to get \((3,-2)\).
But this represents an opposite reciprocal slope than the original! So the lines are perpendicular.
Distance To A line II
Intersecting Circles
Let \(C_1\) and \(C_2\) be two circles in the plane, where the distance between \(d\) their centers is less than or equal to the sum of their radii \(r_1+r_2\). We wish to prove these two circles intersect.
First, lets use a translation isometery to move the plane until \(C_1\) is centered at the origin. Because we know the equation of a circle given its center and radius (Theorem thm-equation-for-euclidean-circles) we can write down an equation for \(C_1\):
\[x^2+y^2=r_1^2\]
Now \(C_2\) was moved to somewhere, and its center is located at some point \(p\). Use another isometry which fixes the origin \(O\), and rotates \(p\) onto the \(x\)-axis.
Rotations about the center of a circle don’t move that circle (Proposition Isometries Fixing the Center Preserve the Circle), so this second isometry doesn’t change \(C_1\). But now we know the center of \(C_2\) lies on the \(x\)-axis, at distance \(d\) from \(O\). This means (by the distance formula Theorem The Euclidean Distance, or the fact that the x-axis is a line Corollary The x-axis is a line) that we know the center of \(C_2\) is at \((d,0)\) in the plane, and so we can write down its equation
\[(x-d)^2+y^2=r_2^2\]
Again, to save ourselves some algebra, we’ll finish the problem by introducing the specific numbers suggested: we’ll say the first circle has radius \(3\), the second has radius \(4\), and the distance between them is \(5\). Given the isometries we’ve used so far we end up with circles with very nice equations:
\[x^2+y^2=3^2\hspace{1cm}(x-5)^2+y^2=4^2\]
To find a pair \((x,y)\) that lies on both circles we need a common solution to these equations. We can find this in many ways, but one of the easiest is to move all the terms of each to one side, then set them equal.
\[x^2+y^2-9=(x-5)^2+y^2-16\]
From here, its just the sort of algebra you’d see in a precalculus course: we expand it out, cancel an \(x^2\) and \(y^2\) from both sides, and solve the resulting equation for \(x\):
\[-9=-10x+25-16\] \[10x=18\implies x=\frac{18}{10}\]
But now, does this \(x\) value really work? To find an actual point of intersection we need that there is a \(y\)-value that goes with it! So, we need to see that we can find a common value of \(y\) such that \((x,y)\) lies on both of our circles
To lie on \(C_1\) we need \(x^2+y^2=r_1^2\), so \(y=\pm\sqrt{r_1^2-x^2}\). Thus, we need that \(r_1^2-x^2\) is positive: otherwise this \(y\) wont exist! To lie on \(C_2\), similarly we need that \(r_2^2-(x-d)^2\) is positive, so we can take the square root and get a \(y\) value.
Plugging in our \(x\) value here, we see that
$\(r_1^2-x^2= 9-(18/10)^2=\frac{144}{25}=\frac{12^2}{5^2}>0\).
But, because we found our \(x\)-value by setting the two equations equal we know that \(r_1^2-x^2=r_2^2-(x-d)^2\) (just cancel the \(y^2\) from both sides). Thus, the quantity needed for the other circle is also positive, and equal to the same value! This tells us the intersection points explicitly:
\[(x,y)=\left(\frac{18}{10},\pm\frac{12}{5}\right)\]
Parabolas
First, we can meausre the distance from a point \(p=(x,y)\) in the plane to a point \(f=(0,h)\) using the distance formula:
\[\begin{align*} \dist((x,y),f)&=\sqrt{(x-0)^2+(y-h)^2}\\ &=\sqrt{x^2+(y-h)^2} \end{align*}\]
If we have a horizontal line with \(L\) with \(y\) coordinate \(-\ell\), we can measure the distance of any point \((x,y)\) in the plane to \(L\) using homework problems 3 and 4. There we learned that the distance to a line is minimized if you reach it via a perpendicular. Since our line is horizontal, perpendiculars are vertical segments, so the closest point to \((x,y)\) is \((x,-\ell)\) with distance
\[\begin{align*}\mathrm{dist}((x,y),L)&=\sqrt{(x-x)^2+(y-(-\ell))^2}\\ &=|y+\ell| \end{align*}\]
The point \((x,y)\) lies on the parabola defined by \(f\) and \(L\) if these distances are equal: that is, if
\[\sqrt{x^2+(y-h)^2}=|y+\ell|\]
We wish to figure out for which values of \(\ell,h\) this describes the standard parabola. First, we can simplify everything by squaring:
\[x^2+(y-h)^2=(y+\ell)^2\]
Expanding this out and collecting like terms, we find
\[\begin{align*}x^2&=(y^2+2\ell y+\ell^2)-(y^2-2hy+h^2)\\ &= 2(\ell-h)y+\ell^2-h^2 \end{align*}\]
To make this into \(y=x^2\) we need that the coefficinet of \(y\) is euqal to \(1\) and the constant term is zero. That is
\[2(\ell+h)=1\hspace{1cm} \ell^2-h^2=0\]
The second equation requires that \(\ell^2=h^2\) so \(\ell=h\) (since both are positive). Plugging this into the first we get \(2(\ell+\ell)=1\) or \(4\ell=1\) so \[\ell=\frac{1}{4}=h\]
Thus, \(y=x^2\) is the parabola with focus \(f=(0,\tfrac{1}{4})\) and directrix the horizontal line at \(y=-\tfrac{1}{4}\).