Odds and Ends

Scientific Factoids

I had ambitious plans for this edition of ‘Odds and Ends’. I was going to come up with a cool little nugget – a science factoid – for every major branch of science – biology, chemistry, geology, physics, astronomy, mathematics.  Because of family and professional obligations, I just didn’t have time to pull it all together, so my compromise is to spread it out over a couple of editions.  This time, it’s biology and mathematics. 
 
Maybe some of you can send me some suggestions for subsequent editions – particularly chemistry or physics.  Everything I’ve come up with so far is either terribly trivial (nuances of the periodic table) or silly (‘the teddybear and the toilet plunger’ trick). Not that the following is much better,,, but might tickle your nerdnik bone.  

Biology

A common sight on the high western desert, the pronghorn ‘antelope’ is not an antelope at all, but a North American ungulate whose closest extant relative is the giraffe. While colloquially known as an antelope, it’s a technically incorrect appellation. Most wildlife biologists just leave it at ‘pronghorn’… and so on a related note, that other ubiquitous western desert resident – the ‘jackalope’ – should more correctly be called the ‘jackahorn’.
 
Pronghorn are ‘wicked’ fast, easily outpacing any living predator within its geographical distribution. While not substantiated by any verifiable research I could find, the pronghorn is often cited as second only to the cheetah in top speed and would leave a cheetah eating dust and sucking wind in any race longer than a quarter mile.

Mathematics

The concept of ‘mathematic elegance’ was lost on me until about half way through my high school geometry class when Mrs. Gore (yes, that was her real name,,, and one of the best teachers I ever had) presented the class with the following proof of the Pythagorean Theorem (see Figure 1 below).  You know,, that oft repeated theorem that, “the sum of the squares of the legs of a right triangle is equal to the square of the hypotenuse” (e.g., The Scarecrow in “The Wizard of Oz”, 1939 to demonstrate the he really had a brain

• 1. Start by constructing a square, each side with length ‘c’.

• 2. Construct an acute angle, α, in the lower left hand corner of the square with the vertex of the angle co-located with the corner of the square.

• 3. Rotate the square clockwise 90 degrees and construct the same angle, α, in the ‘new’ lower left hand corner.

• 4. Repeat Step 3 twice more until all 4 corners have an angle (α) constructed in the corner.

• 5. Extend each of the four (4) angle vectors until they converge at the interior of the large square as illustrated in Figure 1.
  

• 6. Since a right angle, by definition measures 90 degrees, each of the angles adjacent to the α angles has a measure of (90-α) degrees.
  

• 7. Using Area 1 as a reference triangle, and the knowledge that the sum of all three angles of a triangle = 180 degrees, we can show algebraically that the interior angle of the reference triangle is a right angle, i.e., 180 = (90-α) + α + ‘interior angle measure’,,, so ‘interior angle measure’ = 90 degrees.
  

• 8. Therefore, Area 1 is a right triangle with hypotenuse = ‘c’, and we assign the lengths of the other legs ‘a’ and ‘b’ as shown in Figure 1.
  

• 9. Areas 2, 3, and 4 are easily demonstrated to be right triangles congruent to the Area 1 triangle, since they have equivalent measures on two angles and a contained side (α, 90-α, and side ‘c’).
  

• 10. Because of the congruence of the triangles, Area 5 can be shown to be a square with side ‘b-a’ (the difference between the ‘b’ leg of Area 2 and the ‘a’ leg of Area 1, as shown).
  

• 11. The area of the large square = c^2.
  

• 12. The area of the large square also equals the sum of its parts,, i.e, the sum of Areas 1-5.
  

• 13. Area 5 = (b-a)^2.
  

• 14. Areas 1 – 4, each equal ½ ab (the area of a triangle = one half of base x height).
  

• 15. So the sum of Areas 1 – 5 = (b-a)^2 + (4 x ½ ab), which also = c^2 (area of the large square)
  

• 16. Finally,, c^2 = (b-a)^2 + (4 x ½ ab)