 Math Quiz 8

Credit to Jim Goehmann:

(K+1)^2 / (4K(K-1))

And how did you arrive at the answer? The benefit for everyone else reading this is to know the solution process, not just the final answer!

That answer is not correct.

I hadn’t actually figured out if it was correct or not… Regardless, I want to see process!. There are usually a couple of different ways to approach a problem, and seeing someone else’s process can help expand your own thinking I’ll probably take a crack at it later this week. Off the top of my head, we know the equation for a circle. The big circle is pretty well defined, sitting as it does with tangents x=0 and y=0. Then I’d try to define the second circle in relation to the first circle - a circle with the prescribed diameter ratio, a tangent at x=0 and a shared point with the big circle. From there, it’s just a question of finding equation for the hypotenuse that is also a tangent for both circles. That’ll lead you to the x and y intercepts, the length of the hypotenuse, and the length of the base.

I just need to go look up those tangent equations again… Been way too long since I learned them!

I’m happy to see these again!

I got (k+1)^2 / (4 * sqrt(k) * (k-1)), given k > 0 and the diameters != 0

My working is attached below. I mixed up the question, and the equation written is the ratio of base to hypotenuse instead of the other way around. It took me a while to realize some of the geometry the circles provide (and accept the fact that I’d need to use trig), so this is the simplest but not the first solution I found.

Edit: k > 1

Brava! Reps to you (again) Attached is my solution.

Got a chance to talk to Jim again. Lost the sqrt on the k!

Well done Rachel! and thanks for the challenge Ether!

I tried a quick and dirty analytic geometry solution during lunch (essentially equivalent to Rachel’s solution), and made some algebraic mistakes. I’m convinced there’s a far more elegant geometric solution. Along this line (and admittedly working from the known answer), I have worked backwards through Heron’s formula for the area of a triangle and shown that the area of the vertically aligned rectangle bounded by the centers of the two circles is the ratio of the base to the hypotenuse, when the length of the diagonal of that rectangle is defined as unity. This is just too pat an answer not to have a geometric meaning.

Here’s the simplified calculations showing that to be the case:

If you let the radius of the small circle be a, then the radius of the large circle is Ka. The hypotenuse of the tinted rectangle is obviously a(K+1). As this is defined to be length 1, we have that a=1/(K+1). We can easily see that the base is a(K-1), or (K-1)/(K+1). The height can be calculated from the Pythagorean theorem to be 2√K/(K+1). Finally, the area of the rectangle is 2√K(K-1)/(K+1)2, which matches the ratio of base to hypotenuse of the large triangle.

OBTW: image of this rectangle has been posted, and will be linked when available.

Edit: link Though when I awoke this morning I realized that the area of the rectangle was simply sinθ/2 cosθ/2, so I missed by a factor of two and a reciprocal in any case. :sigh:

*Here’s a solution which involves no trigonometry, and no Cartesian geometry. Just simple geometry and some algebra.

It is considerably messier than the trig solution posted earlier.

No_trig solution.pdf (34.2 KB)

No_trig solution.pdf (34.2 KB)

While nibbling on lo mein leftovers at dinner this evening, I finally found a reasonably geometric solution (does not explicitly use double angle formula). It only needs the pythagorean theorem, similar triangles theorem, and some even more basic theorems. Oh - and an isosceles triangle theorem, that if two angles of a triangle have the same value, the sides opposite have the same length.

Note that “long leg” and “short leg” are based on the diagram as presented. In order for the diagram to appear as presented, the apex angle must be acute, implying that the angles labeled φ have a span of less than 45° so this is not an ambiguous definition.

1. Construct perpendiculars and lines as shown in figure which I just uploaded and will link.
2. Note that all angles labeled φ are easily shown to be equal to each other.
3. Define the radius of the small circle as 1. Then, by Ether’s definition, the radius of the large circle is K.
4. Note that the length labeled Q (terminated by open arrows in the diagram) is equal to the ratio of the hypotenuse to the base of the large triangle, by benefit of the apex triangle having a base of length 1 and being similar to the large triangle. That is, Q is that ratio which is to be calculated. [Yes, this is the Q in Q.E.D.]
5. Consider the right triangle with a hypotenuse connecting the centers of the two circles, and legs parallel to the legs of the large triangle. The hypotenuse has length K+1 (being the sum of the two radii 1 and K), and the base has the length K-1, being the difference between these same two radii.
6. We then use the pythagorean theorem to show that the height of that small triangle is 2√K (X2 + (K-1)2 = (K-1)2 ==> X2 = 4K ==> X = 2√K). (not shown on diagram)
7. Next, determine the length of the segment from the center of the small circle to the apex of the large triangle. Because this triangle between these two points and the tangent point of the small circle with the vertical side of the large triangle is similar to the triangle just analyzed, and has a base of 1, its hypotenuse is (K+1)/(K-1).
8. Next, consider the triangle with vertices at the center of the small circle and the two endpoints of segment Q. Because two of its angles are equal, it is an isosceles triangle, with the sides opposite the equal angles having equal lengths. This also means that when the perpendicular bisector to the long side is added as shown, these two triangles are equivalent.
9. Because the sum of the long legs of these two right triangles is (K+1)/(K-1), and the two long legs are equal, each is of length (K+1)/2(K-1).
10. Finally, we note that the right triangle including segment Q as its hypotenuse and angle φ at the apex is similar to the triangle with the two circle centers. Is long leg is length (K+1)/2(K-1), and the ratio of the hypotenuse to the long leg to is (K+1)/2√K. We therefore find that segment Q is of length (K+1)2/(4√K(K-1)).
Q.E.D.

Interesting. I cringed when I saw the trig solution because I thought it was far messier than the Euclidean solution I had sketched out (basically the same one you’ve presented here). In general I find Euclidean geometry to be far more elegant than anything using trig or Cartesian geometry. I guess it’s just a matter of preference.

Cringed is too strong for my reaction, but quite similar. I’d much rather find the geometric path than the analytic geometric path; its much easier to avoid mistakes, as well.

I had actually been rather close to my solution much earlier, but failed to notice that triangle was of the isoceles clan; that was the key.

*I really like Gus’ lo mein approach which eliminates the quadratic equation.

I changed the proof slightly (congruent triangles instead of isosceles) and added color-coded figures to make the proof easier to follow.

*

MathQuiz8Geo.pdf (345 KB)

MathQuiz8Geo.pdf (345 KB)

Ether,
Thanks for the cleanup. I realized during the writeup that I should have labeled some points, but I had already posted the picture.

I also see now what you mean by your no-trig solution being messy. I was wondering why you didn’t do the algebra, then when I tried to solve for DB, I saw why. When I realized that the formula for DB involved taking a square root of a polynomial beginning with 32K4, I really did cringe.

Also, a quick review of geometric proofs of the double angle formula shows that many of them involve an isoceles triangle either explicitly or implicitly. A rather neat one simply involves calculating the area of an isoceles triangle twice (using different sides as the base in each case), and setting the two answers equal to each other.

I did do the algebra… but didn’t post it. I was waiting for someone to comment.

So now that you’ve brought it up, here’s Part B of Math Quiz 8:

Show the algebra steps necessary to convert

ans2=(sqrt((K^2+2*K+1)^2/(K^2-6*K+1)^2-1)+sqrt((K+1)^2/(K-1)^2-1))/(1-(K^2+2*K+1)/(K^2-6*K+1))

to

ans1=(K+1)^2/(4*(K-1)*sqrt(K))

The two expressions above are numerically equivalent in the domain of interest.

*

I can’t get it to work out unless I change the sign on the first term in the numerator (or other equivalent changes, like perhaps changing the signs of both the second term in the numerator and the denominator). Algebra attached.

Edit: - it is only necessary to choose the algebraically negative square root of the first term.

Edit2: This is because K2 - 6K + 1 is negative over the range plotted, and in general over the range 3 +/- √8, or 0.1716 to 5.8284.

Math-Quiz-8B.docx (15.4 KB)

Math-Quiz-8B.docx (15.4 KB)

The square roots in the original expression as given follow the standard convention
and they are positive: if you negate the first term in the numerator of the original expression as given,
it is no longer correct for the domain of K applicable to this problem (1 < K < 3+2√2).

See attached plots.

When you do algebraic manipulations on the expression (or any expression containing squared terms),
you have to be very careful to chose the proper sign of the resulting changed expression.

I tried to give you reps for your substantial contributions to the thread but it wouldn’t let me Yes, this is what I meant by algebraically negative. That is, the square root of the square of K2 - 6K + 1 is not K2 - 6K + 1, but -(K2 - 6K + 1) over the range of the problem.

As the large triangle has a right angle at the lower left, the other angles must be acute. As you noted previously (in a PM as I recall), as the apex angle approaches a right angle, K approaches 3+2√2, the larger root of K2 - 6K + 1.

For those who do not see why this is a maximum value for K, there’s a picture attached with the apex angle being a right angle.

Stipulating a cartesian coordinate system with an origin at the right angle and scaled so that the small circle has radius 1, it is easy to show that the coordinates of point T (the tangent point of the two circles) are (1 + √2/2, -1 - √2/2) from their location on the small circle and also K(1 - √2/2, -1 + √2/2) from their location on the large circle. Solving either the X or Y coordinates of these for K gives a value of (1 + √2/2)/(1 - √2/2), which can be simplified by multiplying by 1 in the form of (1 + √2/2)/(1 + √2/2) to be 3 + 2√2.  To be pedantic for a moment:

The square root of the square of K2 - 6K + 1 is the absolute value of K2 - 6K + 1, which happens to be equal to - (K2 - 6K + 1) for the domain of interest 1 < K < 3+2√2

So ans2 is correct as posted, but if you perform algebraic operations on it to simplify it, you have to careful to assign the correct sign to the resulting changed form.

I think we are saying the same thing using different language.

Many CAS systems handle this by inserting absolute value (||) operators when appropriate if the domain has not been explicitly stated in the problem.