Posts Tagged ‘Geometry’

Monday Math 157

August 4, 2014

Consider a triangle, which we label ∆ABC, with circumcenter O and circumradius R=AO=BO=CO. Let us label the midpoints of the sides as MA, MB and MC, so that MA is the midpoint of BC (the side opposite A), and similarly, so that \overline{AM_{A}}, \overline{BM_{B}} and \overline{CM_{CA}} are the medians. Then \overline{OM_{A}}, <\overline{OM_{B}} and \overline{OM_{C}} are segments of the perpendicular bisectors of the sides. Let us also label the feet of the altitudes from A, B and C as HA, HB and HC, respectively, and let H be the orthocenter (the intersection of the altitudes \overline{AH_{A}}, \overline{BH_{B}} and \overline{CH_{C}}).


Let us construct the point D on the circumcircle diametrically opposed to A; that is to say, the point D such that AD is a diameter of the circumcenter. Then AD=2R, and O is the midpoint of AD.

Now, by Thales’ theorem, ∠ABD and ∠ACD are both right angles. Now, since CD and the altitude \overline{BH_{B}} are both perpendicular to AC, they are parallel to each other. Similarly the altitude \overline{CH_{C}} and segment BD are parallel, both being perpendicular to AB. Thus, the quadrilateral BDCH is a parallelogram.


Since the diagonals of a parallelogram bisect each other, we see that the midpoint MA of BC is also the midpoint of HD.

Now, let PA be the midpoint of the segment AH. Then \overline{P_{A}M_{A}} is a midline of the triangle ∆AHD, and by the triangle midline theorem, \stackrel{\longleftrightarrow}{P_{A}M_{A}}\parallel\stackrel{\longleftrightarrow}{AD} and PAMAAD=R.

Now, let N be the intersection of \overline{P_{A}M_{A}} and HO. By the midline-median bisection theorem proven in this post, we see that, as HO is the median of ∆AHD that crosses midline \overline{P_{A}M_{A}}, N is the midpoint of both \overline{P_{A}M_{A}} and HO. Thus, NMA=NPAPAMAR.

Now, consider the quadrilateral HOMAHA. Since \overline{HH_{A}} and \overline{OM_{A}} are both perpendicular to BC, HOMAHA is a right trapezoid. Letting QA be the midpoint of \overline{H_{A}M_{A}}, we see then that NQ is the median (or midline) of trapezoid HOMAHA.


By the first of the three items proven here, we see that \overline{NQ}\parallel\overline{HH_{A}}\parallel\overline{OM_{A}}, and so \overline{NQ}\perp\overline{H_{A}M_{A}}. Thus, we see that \stackrel{\longleftrightarrow}{NQ} is the perpendicular bisector of \overline{H_{A}M_{A}}, and so, by the perpendicular bisector theorem, NHA=NMA, and so

Constructing diameter BE of the circumcircle gives us parallelogram CEAH, by a similar argument as above. Letting PB be the midpoint of the segment BH, analogous reasoning to the above shows that NHB=NMB=NPBR as well. Lastly, diameter CF, and midpoint PC of CH gives, by similar proof, that NHC=NMC=NPCR. Thus, the nine points MA, MB, MC, HA, HB, HC, PA, PB and MC are all equidistant from N.

Therefore, we see that for any triangle, the midpoints of the sides, the feet of the altitudes, and the midpoints of the segments connecting the vertices to the orthocenter all lie on a single circle, which, for this reason, is usually known as the nine-point circle, and its center N the nine-point center. We also see that the radius of the the nine-point circle is half the radius of the circumcircle, and the nine-point center is the midpoint of the segment connecting the circumcenter and orthocenter. This latter tells us that for any non-equilateral triangle, the nine-point center lies on the Euler line (for an equilateral triangle, it coincides with the circumcenter, orthocenter, and centroid).



Monday Math 154

July 14, 2014

Continuing the series on triangle centers, let us consider ∆ABC, with circumcenter O and centroid G. Construct the line segment OG, and extend it out from G to the point H such that GH=2OG.

Next, construct the median from vertex A to the midpoint M of side BC. Then G lies on AM, with AG=2GM, as we showed here. Recalling that the circumcenter is the intersection of the perpendicular bisectors of the sides, we see that OMBC.


Now, since AG=2GM, GH=2OG, and ∠AGH≅∠MGO, we see (by the SAS similarity condition) that ∆AGH~∆MGO. And since these triangles are similar, corresponding angles ∠HAG and ∠OMG are congruent. However, these are alternate interior angles for lines \stackrel{\longleftrightarrow}{AH} and \stackrel{\longleftrightarrow}{OM} cut by transversal \stackrel{\longleftrightarrow}{AM}, and therefore \stackrel{\longleftrightarrow}{AH}\parallel\stackrel{\longleftrightarrow}{OM}. And since OMBC, we see \stackrel{\longleftrightarrow}{AH}\perp\overline{BC}, and \stackrel{\longleftrightarrow}{AH} is the triangle altitude from A to BC.

Analogous constructions show that H must also be on the triangle altitudes from B and C:


Thus, we see that H is the orthocenter of ∆ABC. So, we see that for any non-equilateral triangle, the circumcenter O, centroid G and orthocenter H are collinear, with GH=2OG; the line through these triangle centers is known as the Euler line of the triangle. (For an equilateral triangle, O, G and H are all the same point.)

Monday Math 152

June 30, 2014

Find the point in the interior of a triangle for which the product of the distances from that point to the sides of the triangle is maximized.


Monday Math 150

June 16, 2014

1) That all three medians of a triangle intersect at a single point (the centroid of the triangle), and that this point divides the medians into segments with a 2:1 length ratio.


2) That the six smaller triangles into which a triangle is divided by its medians have equal area.

Monday Math 147

December 20, 2010

Given two non-zero complex numbers z1 and z2 such that \left|z_1+z_2\right|=\left|z_1-z_2\right|, show that the arguments of z1 and z2 differ by π/2.

Monday Math 141

November 1, 2010

Let us combine the results from the previous two weeks (here and here). We found that for a regular n-gon with unit sides, the diagonals have lengths d_k=\frac{\sin\frac{k\pi}n}{\sin\frac{\pi}n}, k=2,3,…,n-2, with d1=dn-1=1 the length of the sides.
Now, then, we consider the product of two diagonal lengths, dkdm. From the above, this is
We found here that the numerator product is
Now, if km, then km+2i-1 is a positive integer for all i=1,2,…,m; similarly, if k+mn, then the largest value of km+2i-1 in the sum, k+m-1, is then an integer less than n, and thus the term \frac{\sin\frac{(k-m+2i-1)\pi}n}{\sin\frac{\pi}n} is equal to a diagonal (or side) d_{k-m+2i-1} for all i in the sum, so
d_kd_m=\sum_{i=1}^{m}d_{k-m+2i-1}, with 2≤mkn-2, k+mn. This is the “diagonal product formula” named by Peter Steinbach, who outlined a proof using the n-gon formed by the complex n-roots of unity (compare this problem, for example).
Now, we can generalize the formula first by exchanging k and m to see that for mk,
we combine to form
which gives us the formula for any 2≤kn-2, 2≤mn-2, k+mn.
For k+mn, we see that one or both of k and m must be greater than \frac{n}2. However, recall that dk=dnk. Thus, if d_k>\frac{n}2, then d_{n-k}<\frac{n}2, and similarly for m; thus, we see we can pick the smaller of k and nk; the latter gives the same terms in the sum, just in opposite order, so only the limit matters. Doing the same with m, we thus find the formula for all k and m in the range 2 to n-2:

Now, let us look at a few examples. The smallest n to give us a diagonal is n=4, the square. For square of unit side, the length of the diagonal is \sqrt2, and so the only diagonal product is
d_2^2=\sum_{i=1}^{2}d_{2i-1}=d_1+d_3, which is
For the pentagon, the diagonals have length \phi=\frac{1+\sqrt5}2, the golden ratio, so the only unique diagonal product, since d_2^2=d_2d_3=d_3^2, is
d_2^2=\sum_{i=1}^{2}d_{2i-1}=d_1+d_3, which is
a classic equation for the golden ratio.
For the hexagon, we have d_2=d_4=\sqrt{3} and d_3=2.
Thus we have three unique products
d_2^2=\sum_{i=1}^{2}d_{2i-1}=d_1+d_3, which is
d_2d_3=\sum_{i=1}^{2}d_{2i}=d_2+d_4, which is
d_3^2=\sum_{i=1}^{3}d_{2i-1}=d_1+d_3+d_5, which is

Further, this can be used to prove some interesting relations of the (non-constructible) diagonal lengths of the heptagon, and thus the sides of the heptagonal triangle. Letting the heptagon have unit side, labelling the shorter and longer diagonal lengths as a=\frac{\sin\frac{2\pi}7}{\sin\frac{\pi}7} and b=\frac{\sin\frac{3\pi}7}{\sin\frac{\pi}7} respectively, then our diagonal product formula tells us (with d_1=d_6=1, d_2=d_5=a, and d_3=d_4=b):
d_2^2=\sum_{i=1}^{2}d_{2i-1}=d_1+d_3, and thus
d_2d_3=\sum_{i=1}^{2}d_{2i}=d_2+d_4, and thus
d_3^2=\sum_{i=1}^{3}d_{2i-1}=d_1+d_3+d_5, which is
Via simple algebra on these equations, we can find formula for the quotients of these numbers; dividing ab=a+b by a or b and solving for the quotient left, we see
\frac{b}{a}=b-1 and \frac{a}{b}=a-1;
and dividing by the product ab gives
while further algebra with these formulas lets us find the reciprocals of a and b as linear combinations of a, b, and 1:

These relations appear, for example, in the substitution rules for Danzer’s 7-fold quasiperiodic tiling and Maloney’s 7-fold quasiperiodic tiling

Monday Math 140

October 25, 2010

What are the lengths of the diagonals of a regular n-sided polygon with sides of unit length?

Monday Math 136

September 20, 2010

Here is a very clever geometry proof I once saw. I do not recall where it was I encountered it, but It sticks out in my memory for its creative simplicity; I take no credit for it.

Statement: suppose we have a box (right rectangular prism), which is divided completely into a finite number of smaller boxes, not necessarily the same size or shape. If for each of these smaller boxes, at least one of the edges (dimensions) is of integer length, then at least one edge of the overall box must be of integer length.

Monday Math 120

May 24, 2010

Given a (non-degenerate) conic section, other than a parabola, draw two parallel lines, each of which intersects the conic at two points. Next, find the midpoints of these two parallel chords. Then the center of the conic section lies on the line connecting these two midpoints. This can be used to find the center (and thus the axis and focus/foci) of any conic with only straightedge and compass.

Here, we will give an analytic proof. Let our conic be written as . If p and q are both positive, we have an ellipse (or a circle, if p=q); if they differ in sign, we have a hyperbola. In all cases, the center is at the origin. Given the line , we substitute to find the intersections:

If the line has two intersections with the conic, then the discriminant of the above quadratic is positive. The x-coordinate of the midpoint is the average of the solutions to the above quadratic; considering the quadratic formula, this is
since the midpoint lies on the line, the y-coordinate is
The line connecting this to the origin has slope
Note that this is independent of the intercept b of the intersecting line. Thus, a parallel line will give a midpoint that gives the same slope, and thus lies on the same line through the center.

For the parabola, we note that it can be considered the limiting case of an ellipse, or hyperbola, as the eccentricity approaches 1 while the center goes to infinity; hence, a line through the midpoints of paralell chords, and thus through the center, will in the limiting case of the parabola will be parallel to the axis of the parabola (the path of the center as it moves toward infinity).

Monday Math 48

December 1, 2008

Previously, we noted that the hyperboloid of one sheet is one of only three surfaces to be doubly ruled; that is, to be able to be swept out by a moving line in more than one distinct way.
A brief bit of thought should give a second example of a doubly ruled surface, the most trivial of the three: the plane. Not only does the plane admit two distinct rulings (draw a grid on the plane for an example of two rulings), but infinitely many: rotate a ruling of a plane by any angle about an axis normal to the plane and you’ve generated a new, distinct ruling. In fact, the plane is the only n-ruled surface for n>2.

Now, how about the third surface?

Let us consider a line to be swept to form one ruling, and let us choose the line in one particular position to be the y-axis of our coordinate system. Now, let us choose an x-axis, and let us sweep our y-axis line such that all our lines in this first ruling are of constant x coordinate: that is, our lines are of the form x=x0, z=m(x0)y+b(x0), so we have surface z=m(xy+b(x), with continuous functions m(x) and b(x) with m(0)=b(0)=0.

Now, we want m(x) and b(x) such as to admit a second ruling. First, let us consider z=0, which occurs at (note that the singularity at x=0 is removable). Here, we can choose our origin (and thus x-axis) so that it is the point where this curve crosses the y-axis: namely, so that , and b(x) goes to zero faster than m(x) as x→0.
To make this a line, we need , so that b(x)=-px·m(x), with some constant p.

Thus, our surface is z=m(x)·(ypx). Now, we want a family of lines swept from the line y=px, z=0, which also gives this surface. Let us examine the behavior of z when y=px+q, for q≠0 a constant. Then we have z=m(x)·(px+qpx)=q·m(x). We see immediately that if m(x) is a linear function of x, m(x)=ax (as m(0)=0), then these slices of our surface are automatically lines: z=aqx, y=px+q. Then our surface is just z=ax(ypx)=axyapx2. The p=0 case, z=axy, is a hyperbolic paraboloid. The p≠0 case is just this under a shear transformation, and is thus a hyperbolic paraboloid as well (as a shear can be formed from the composition of rotations and a non-uniform scaling). For each, there is an affine transformation that can map to the general hyperbolic paraboloid , and so the (general) hyperbolic paraboloid is our third doubly ruled surface.