Find a non-summation expression for the value of the sum .
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 , and are the medians. Then , < and 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 , and ).
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 are both perpendicular to AC, they are parallel to each other. Similarly the altitude 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 is a midline of the triangle ∆AHD, and by the triangle midline theorem, and PAMA=½AD=R.
Now, let N be the intersection of 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 , N is the midpoint of both and HO. Thus, NMA=NPA=½PAMA=½R.
Now, consider the quadrilateral HOMAHA. Since and are both perpendicular to BC, HOMAHA is a right trapezoid. Letting QA be the midpoint of , 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 , and so . Thus, we see that is the perpendicular bisector of , 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=NPB=½R as well. Lastly, diameter CF, and midpoint PC of CH gives, by similar proof, that NHC=NMC=NPC=½R. 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).
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:
- the median of a trapezoid is parallel to the bases;
- the length of the median is half the sum of the lengths of the bases;
- the midpoints of the diagonals of a trapezoid also lie on its midline.
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 and cut by transversal , we see that . 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 . And since opposite sides of a parallelogram have equal length, DF=BC, and so DE=½DF=½BC.
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
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 and cut by transversal , 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 DE=½DF=½BC.
Let MA, MB and MC be the midpoints of sides BC, AC and AB, respectively, of ∆ABC. Thus, is a midline of ∆ABC, and a median. Let P be the point where they intersect.
By the midline theorem, and MBMC=½BC.
This means, then, that , and so by our second theorem above with regards to ∆ABMA, we see that P must be the midpoint of , and MCP=½BMA.
Similarly, our second theorem applied to triangle AMAC establishes that MBP=½CMA. But BMA=CMA, and so MBP=MCP, and P is the midpoint of as well.
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 OM⊥BC.
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 and cut by transversal , and therefore . And since OM⊥BC, we see , and 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.)
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.
•”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
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“
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 Economist “Why 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.