Math Quiz 8

[Ether]

Now that build season is over, it's that time again.

[mastachyra]

Credit to Jim Goehmann:

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

[Jon Stratis]

Originally Posted by mastachyra

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!

[Ether]

Originally Posted by mastachyra

Credit to Jim Goehmann:

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

Originally Posted by Jon Stratis

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.

[Jon Stratis]

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!

[Rachel Lim]

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

[Ether]

Originally Posted by Rachel Lim

I'm happy to see these again!

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

Brava! Reps to you (again)

Attached is my solution.

[mastachyra]

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

Well done Rachel! and thanks for the challenge Ether!

[GeeTwo]

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:

[Ether]

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.

[Karthik]

Originally Posted by Ether

It is considerably messier than the trig solution posted earlier.

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.

[GeeTwo]

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.

[Ether]

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.