Angles without the angles...
I saw this in a Martin Gardner book a while ago. Shown in the figure are 3 squares, and 3 angles in the bottom left. Prove, without any trigonometry (or equivalent), that the sum of the two smaller angles equals the largest.
~ Grenade
posted by littlecow at 4/07/2006 07:51:00 AM
5 Comments:
Is pythogoras theorem allowed? [it ought to be as it can be proved drawing pictures without resorting to trigonometry]
In figure http://i83.photobucket.com/albums/j306/bbalasub/P1010047.jpg,
the problem is to show that the angles marked by "x" are equal.
Triangle OPQ has sides 3,1 and sqrt(10) units. Triangle OCD has dimensions 3 (OA,AB,BC), 1 and sqrt(10). The triangles are similar. Therefore, the "x" angles are equal.
Yes,I thought so too,but what would that construction be?
Probably the nomenclature
HGFE
ABCD
can be used to explain that.
The solution is very simple if you stack two rows of three squares each on top of this row of three squares. Now you have a square with 9 equal mutually exclusive squares covering its area. The largest angle can be extended to be the diagonal of the larger square, where as the smaller angles hit the 1/3 and 1/2 marks on the right verticle side of the larger square. (So, trigonometric proof is as trivial as arc tan 1/3 + arc tan 1/2 = arc tan 1). For the geometric proof, let me name the points on the big square. Let the bottom left point be (0,0) and top right be (3,3). Thus, the small angle connects (0,0) and (3,1), middle angle connects (0,0) and (2,1) and (3,1.5), and largest angle connects (0,0), (1,1), (2,2), and (3,3). Call (0,0) A, (3,3) B, (3,1) C, (3,1.5) D, (3,0) E, (0,3) F. Little angle is CAE, middle angle is DAE, and large angle is BAE and is 45 degrees.
Proof:
Where FE intersects AB call it G (center of big square), where FE intersects AD call it H, where FE intersects AC call it I. All we have to do is to prove that GE is trisected at H. This is obviously true if you join (1,0) with (3,2) as it passes through H. In triangle AEB, sides AE and EB are trisected at points (1,0) and (3,2) respectively, hence line joining (1,0) with (3,2) is parallel to line AGB and it trisects the line GE at H (since it is perpendicular to line AGB). Thus, triangles AGH and AEC are similar making angle GAH = angle EAC. And angle GAH + HAE = Angle GAE. Or, angle EAC + angle DAE = angle BAE. Or, small angle + medium angle = large angle.
QED.
Very cool design! Useful information. Go on! » » »
Post a Comment
<< Home