## Monday Math 156

July 28, 2014

For a trapezoid, the median (also known as the midline or mid-segment) is the segment connecting the midpoints of the legs of the trapezoid. This post presents a proof that:

1. the median of a trapezoid is parallel to the bases;
2. the length of the median is half the sum of the lengths of the bases;
3. the midpoints of the diagonals of a trapezoid also lie on its midline.

## Monday Math 155

July 21, 2014

The triangle midline theorem, also called the midsegment theorem, states that the line segment connecting the midpoints of two sides of a triangle is parallel to and half the length of the third side. This may be trivially proven via triangle similarity (SAS similarity condition) and the corresponding angles postulate. More interesting, however, is to prove it using triangle congruence.

Let D and E be the midpoints of sides AB and AC, respectively, of ∆ABC. Let us extend segment DE past E to point F such that DE=EF, and let us draw CF.

Since DE=EF, AE=EC, and vertical angles ∠AED and ∠CEF are congruent, we see by the SAS condition that ∆ADE≅∆CFE. Thus, CF=AD=BD. Also, ∠FCE≅∠DAE; but since these are alternate interior angles for lines $\stackrel{\longleftrightarrow}{AD}$ and $\stackrel{\longleftrightarrow}{CF}$ cut by transversal $\stackrel{\longleftrightarrow}{AC}$, we see that $\stackrel{\longleftrightarrow}{AD}\parallel\stackrel{\longleftrightarrow}{CF}$. But then the quadilateral BCFD has a pair of opposite sides, BD and CF, which are of equal length and parallel, so it is therefore a parallelogram, and so $\stackrel{\longleftrightarrow}{DE}\parallel\stackrel{\longleftrightarrow}{BC}$. And since opposite sides of a parallelogram have equal length, DF=BC, and so DEDFBC.

We can also prove a similar theorem: that the line through the midpoint of one side of a triangle parallel to a second side of the triangle bisects the third side, and that the segment of that line inside the triangle is one-half as long as the parallel side. As with the midpoint theorem, it is trivial to prove with triangle similarity (this time the AA condition) and the corresponding angles postulate. So, instead, we use a similar construction as above.

Let D be the midpoint of side AB of ∆ABC, and let E be the point where the line through D parallel to BC intersects AC. Construct the line through C parallel to AB, and let F be the point where it intersects $\stackrel{\longleftrightarrow}{DE}$

Then BCFD is a parallelogram, and since opposite sides of a parallelogram have the same length, BD=CF and BC=DF. And so CF=BD=AD. And since ∠ADE and ∠CFE are alternate interior angles for lines $\stackrel{\longleftrightarrow}{AD}$ and $\stackrel{\longleftrightarrow}{CF}$ cut by transversal $\stackrel{\longleftrightarrow}{DF}$, they are congruent. Similarly, ∠DAE≅∠CFE, and so, by the ASA condition, ∆ADE≅∆CFE. Thus, AE=EC, and E is thus the midpoint of AC. We see also that DE=EF, and since DF=BC, thus DEDFBC.

Using both of these theorems together, we can prove a third: that a midline (mid-segment) of a triangle and the triangle median that intersects it bisect each other.

Let MA, MB and MC be the midpoints of sides BC, AC and AB, respectively, of ∆ABC. Thus, $\overline{M_{B}M_{C}}$ is a midline of ∆ABC, and $\overline{AM_{A}}$ a median. Let P be the point where they intersect.

By the midline theorem, $\stackrel{\longleftrightarrow}{M_{B}M_{C}}\parallel\stackrel{\longleftrightarrow}{BC}$ and MBMCBC.

This means, then, that $\stackrel{\longleftrightarrow}{M_{C}P}\parallel\stackrel{\longleftrightarrow}{BM_{A}}$, and so by our second theorem above with regards to ∆ABMA, we see that P must be the midpoint of $\overline{AM_{A}}$, and MCPBMA.

Similarly, our second theorem applied to triangle AMAC establishes that MBPCMA. But BMA=CMA, and so MBP=MCP, and P is the midpoint of $\overline{M_{B}M_{C}}$ as well.

## 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 153

July 7, 2014

A few weeks ago, I demonstrated a proof that all three medians of any triangle are concurrent; in other words, that the centroid exists.

Now, how about demonstrating that the circumcenter and orthocenter exist; that is to say, showing that the three perpendicular bisectors of the sides are concurrent, and that the three altitudes are concurrent, for any triangle.

## Some links (and quotes) for the 4th of July

July 4, 2014

•”Strictures Upon the Declaration of the Congress at Philadelphia.
Thomas Hutchinson’s anonymously published rebuttal to the Declaration of Independence (and fisking of its charges against George III).

William Bernstein on the Boston Tea Party in this EconTalk podcast (circa 25 minutes in). The takeaway: it was essentially the “first anti-globalization riot”, with smugglers like Samuel Adams protesting a reduction in taxes that was undercutting their business.

Related: Historynet’s Debunking Boston Tea Party Myths
An exerpt:

Resistance leaders also launched a new wave of negative propaganda that played to anti-foreign sentiments: Tea from the East India Company was packed tightly in chests by the stomping of barefoot Chinese and was infested with Chinese fleas. In turn, a vast number of colonists vowed to protect American business from foreign competition, even if that business was smuggling. Beware of products from China, buy America, wage war on drugs, down with corporations—all these messages, as well as their better-known cousin, no taxation without representation—amplified the response to Parliament’s Tea Act of 1773.

•”American Loyalists“: Buried History of the American Revolution.

•Handle: “Suppressing Tories

•A quote:

Sir, As the Committee of Safety is not sitting, I take the Liberty to enclose you a Copy of the Proclamation issued by Lord Dunmore; the Design and Tendency of which you will observe, is fatal to the publick Safety. An early and unremitting Attention to the Government of the SLAVES may, I hope, counteract this dangerous Attempt. Constant, and well directed Patrols, seem indispensably necessary.

Patrick Henry, in a pamphlet written to county lieutenants throughout Virginia concerning the Earl of Dunmore’s proclamation offering emancipation to slaves of Patriots who escaped and joined the Royal forces. (See also Lord Dunmore’s Ethiopian Regiment).

•Tangentially related, particularly the comments threads: The EconomistWhy the first world war wasn’t really” on the Seven Years’ War — and its American theater, the “French and Indian War“/”La guerre de la Conquête” — as the actual first world war. (See also the Battle of Jumonville Glen.)

Update: Nick B. Steves has created a permanent page for “Strictures upon the Declaration of Independence” over at The Reactivity Place.

## 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 151

June 23, 2014

Find the values of x,y,z≥0 that maximize the value of f(x,y,z)=xyz, subject to the constraint ax+by+cz=d, where a, b, c and d are positive real numbers.
solution:

## Monday Math 150

June 16, 2014

Prove:
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.

and

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

## US Supreme Court calls almost one quarter of blacks retards

June 12, 2014

In the recent Hall v. Florida decision, the US Supreme Court (besides demonstrating a limited understanding of statistics; but what do you expect, they went to law school) has effectively raised the IQ threshold with regards to the level of mental handicap to bar execution to 75. However, when one consideres that the median black American IQ is 85, with standard deviation 13.5 (here), this cutoff has a z-score of (75-85)/13.5≈0.74, which corresponds to a percentile of 23%. Thus, SCOTUS has effectively called almost a quarter of American blacks retarded.

## Another thought-provoker suitable for a t-shirt

June 7, 2014

The opposite of “discriminate” is “indiscriminate”.