POTW Returns!

Do you prefer math problems to chess problems? Well then today's your lucky day! For today is the day that Problem Of the Week returns! Our theme this semester is: Geometry. Euclidean geometry to be exact.

There will be ten problems during the term. The first five are meant to be relatively simple, while the second five will be more challenging. So have a look, see what you can make of it, and feel free to leave your solutions in the comments.

More like this

I'll assume from the drawing that AC and CD (both extended) are diameters. Otherwise, there's insufficient information.

Since ABCD is a rectangle, the two diagonals are equal (this follows from SAS triangle congruence); so, AD = BC. Since BC is a radius, AD = 5.

By Michael Kelsey (not verified) on 25 Jan 2015 #permalink

You can take it as a general principle that the diagrams are not deceptive. Things that look like diameters are diameters, things that look perpendicular are perpendicular, and so on.

My first thought was 'damn I do not have enoug information for this, there must be some obscure rule I am forgetting, or perhaps the clue is that it is a rectangle", but then I came to the same conclusion as #1.

I always expect I have to look at tan, cos and sin, and remember the relationships between them. We had about a 100 of those problems to do as homework in high school.

By Soren Kongstad (not verified) on 25 Jan 2015 #permalink

I always expect I have to look at tan, cos and sin, and remember the relationships between them

There was zero need for that here. Instead of trying to think of the rectangle depicted, you merely need to (a) realize that all possible rectangles in the space would have the same diagonal length, and (b) then think of the two "extreme" rectangles where height approaches 5 and width approaches 0 and vice versa; in those cases diagonal = radius.

As it happens, back in 2007 I did a semester where the POTW theme was trigonometry. Click here. That was one of the first terms where I did POTW, though, and I notice I got a bit lazy, since I stopped posting solutions after the third problem. I've always been meaning to go back and fix that, but since I no longer recall the source of some of the problems, I will have to try to solve them myself.

For this term, though, there will be no trigonometry required. Just basic Euclidean geometry: area formulas, Pythagorean theorem, triangle congruence theorems, some basic circle stuff, that sort of thing. But you can make some hard problems with those materials! As I mentioned in the post, the first five are meant to be pretty basic, while the last five are more challenging.

Jason Rosenhouse wrote (#2, January 26, 2015):
> You can take it as a general principle that the diagrams are not deceptive. […]

Then are perhaps the problem statements meant to be deceptive instead? …

After all, the statement of "Problem One, 2015",
http://educ.jmu.edu/~rosenhjd/POTW/Spring15/POTW1S15.pdf
says that

Figure $latex A B C D$ is a rectangle [with] segment $latexA D$

(where it is not spelled out explicitly whether the "segment $latex A D$" is supposed to be a side or a diagonal of the rectangle under consideration)

while the corresponding diagram seems to present rather the rectangle "$latex A B D C$"

(i.e. following to the convention of labeling the vertices alphabetically in the order in which they are consecutively joined by sides; as illustrated for instance with the http://en.wikipedia.org/wiki/British_flag_theorem ).

By Frank Wappler (not verified) on 26 Jan 2015 #permalink

Hang on. I mean, obviously the answer is 5, since it's going to be equal to BC, which is a radius of the circle.

But are we expected to prove that? I presume we are. The proof required would be that the diagonals of a rectangle are of equal length.

Here's my proof. Ignore the circle and just consider the rectangle ABDC. Because it is a rectangle, we know that AC = BD and AB = CD, and that angle BAC = ACD = 90 degrees (also by definition).

Construct the triangles BAC and ACD. The former has the diagonal BC; the latter has the diagonal AD. By the Side Angle Side axiom, since BA = CD, AC = AC, and the included angles BAC = ACD, the triangles BAC and ACD are congruent. Thus, BC = AD.

Since we know BC = 5 (as it is a radius of the circle), it follows that AD = 5.

Or if you prefer theorems, we know that the hypotenuses are equal to sqrt(BA * BA + AC * AC) and sqrt(CD * CD + AC * AC), which (since BA = CD) are therefore equal to each other, via Pythagorus.

Generally I require that, to be eligible for the prize, students must provide a correct answer with a brief explanation. However, they are allowed to make free use of any standard geometrical theorems in providing their explanation. That the diagonals of a rectangle are the same length is a standard theorem, so students can simply refer to it, just like they can use the Pythagorean theorem without proving it first.

As it happens, one of my students came up with a more elaborate solution based on seeing the center of the circle as the origin of a coordinate system. If we say that point B has coordinates (x,y), then we can work out the coordinates of A and D in terms of x and apply the distance formula. The x's will cancel and the answer will appear. Clever, but definitely doing it the hard way.

The coordinate method does have the feature that it basically converts a geometry problem into algebra. Algebraic proofs are often easier to construct. I wouldn't have thought of doing that in this case, but it's a valid technique for harder problems.

A trick I learned in my high school geometry class (a course I loved) was to draw another line, dashed to indicate it wasn't in the original statement. It had a special name which I can't recall (fiduciary line? extensory line?). In this case I draw a line from the center to the right upper corner of the rectangle, which is a radius long, then use side-angle-side to prove its triangles are the same as the ones formed by the other diagonal.

Gazza,

I agree with you. I am not sure why you'd want to use the algebraic method for this one when the geometry is so easy. However, you could even add some trigonometry if you really wanted to. The coordinates of point B (assuming C is the origin and letting A=the measure of angle BCD) would be (rcosA, rsinA). The distance BC would then simply be
sqrt [r^2*(sin^2A + cos^2A)], which is easily reduced to r=5.

BTW, I'm not sure that it's true that algebraic proofs are inherently easier to construct. It may well be that we find them so because of the way that mathematics is taught today. Of course, finding a working definition for the inherent ease of construction of a proof might itself be difficult. In any case, even using an intuitive understanding of difficulty of construction, it's pretty clear that the question of whether a geometrical or an equivalent algebraic proof would be easier to construct is dependent on the specific problem. For this problem, it's pretty clear that the geometric method is easier, for instance. The fact that for most problem we find algebra easier is probably based on the fact that modern mathematical teaching focuses more on numbers and arithmetic than it does on geometrical relationships. For instance, if it were possible to time-transport an ancient Greek to the present day and educate him or her in modern analytical geometry, he or she would still probably find that geometric methods were easier to use, simply based on the fact that his or her mathematical background was founded on geometry rather than algebra.

For anyone interested, Jason has POTW2 posted. A bit harder than 1, but still not too difficult.