Math Quiz 8
 

Re: Math Quiz 8
 

Credit to Jim Goehmann:

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

Re: Math Quiz 8
 

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!

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That answer is not correct.

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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!

Re: Math Quiz 8
 

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

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Brava! Reps to you (again):)

Attached is my solution.

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

Well done Rachel! and thanks for the challenge Ether!

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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:

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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.

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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.

Re: Math Quiz 8
 

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.

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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.

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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.

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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.

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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
to

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

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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.

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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 :(

Re: Math Quiz 8
 

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.

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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.

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Agreed.

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Chatting with a co-worker about this problem on Friday afternoon, we came up with another method (slightly more algebraic) to determine the length of the line segment from the center of a circle to the apex of the triangle, and came up with another quiz.

Alternate calculation: From the center of the circle of radius 1, a series of infinite circles can be inscribed in the circle/angle of radius 1/K, 1/K², 1/K³, etc. The sum of the diameters of this infinite series of circles is 2/(K-1). When added to the radius of the larger circle, you get (K-1+2)/(K-1) or (K+1)/(K-1).

The follow-on quiz is about the golden ratio. Geometrically speaking, a golden rectangle is a rectangle which is similar to the rectangle created when a square is appended on its long side, or equivalently when a square is excised on the short side.
That is, in the diagram uploaded, given that AB=CD, AC:AB = AB:AD.

The quiz:

1) Develop a geometric construction which produces a golden rectangle (simple).
2) prove geometrically (that is, not using any analytic tools, including the quadratic equation or even the pythagorean theorem) that the construction developed in #1 (or another valid geometric construction) produces rectangles meeting the description above.

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A hint on constructing a golden rectangle:

The definition of the rectangle implies that if the ratio of the longer side to the shorter side is named Ф, the following equation holds:

Ф² = Ф + 1

Using the quadratic formula and selecting the positive root yields:

Ф = (1 + √5) / 2

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OK, no takers on the construction, so here it is, along with an analytic proof.

Construction:

• Construct square ABDC, extending sides AB and CD as shown.
• Bisect segment AB to find point E
• Draw circle centered at E with radius ED. Intersection with line AB is point F.
• Draw perpendicular to AF at F; intersection with CD is G

Tedious analytical proof that rectangles BFGD and AFGC are both golden rectangles:

• BFGD and AFGC differ by a square of their common length.
• Define λ to be the length of EB.
• By construction, BD = 2 λ
• By construction, EBD is a right angle
• By the Pythagorean theorem hypotenuse ED = √((2λ)² + λ²) = √5 λ.
• As EF and ED are radii of the same circle, EF = √5 λ.
• BF = (√5-1) λ, AF = (√5+1)λ.
• AF/BD = (√5+1)λ/2λ = (√5+1)/2
• BD/BF = 2λ/(√5-1)λ = 2(√5+1) / (√5-1)(√5+1) = 2(√5+1) /(5-1) = 2(√5+1)/4 = (√5+1)/2
• As AF/BD = BD/BF, BFGD and AFGC are similar.
• As BFGD and AFGC are similar and differ by a square on their common side, they are both golden rectangles.

There's still the matter of a pure geometric proof of this or another construction. There is at least one much more elegant than the analytic proof above, based on essentially this same construction.

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To both bump this thread and provide a hint, here's a proof of Thale's Theorem, which I used in my golden rectangle proof. Thale's Theorem states that an angle inscribed on a circle which intersects a diameter of the circle is a right angle. It is a special case of the general inscribed angle theorem, which states that an angle inscribed on a circle spans an arc of the circle which is twice the angle.

OBTW,  is supposed to indicate an angle. It looks like CD has put in a different symbol.

• Begin with BAD inscribed on circle centered at C across diameter BD.
• Draw segment CA.
• The triangle BAD has angles:
• CBA + CAB + CAD +CDA
• Because CB, CA, and CD are all radii of circle C, they have the same lengths.
• Triangles BCA and ACD are isosceles. This implies that:
• CAB = CBA
• CAD = CDA
• The angles of triangle BAD therefore add up to: 2 CAB + 2 CAD.
• BAD =CAB + CAD, so the sum of the angles of triangle BAD is therefore 2 BAD.
• The sum of a triangle's angles is two right angles, therefore BAD is a right angle. Q.E.D.

Additionally, Jason and I have not let Ether's original question go; we spend a few minutes each day looking for alternate solutions. I have a derivation for the original quiz which does not require the Pythagorean Theorem, just similar triangles and a few odd (but easily proven) circle/angle theorems. As a clue, let me note that the vertical distance between the centers of the two circles equals the geometric mean of the diameters of the two circles. The constructions in my derivation are anything but obvious.

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Rather than leaving this hanging, here's the geometric proof of a golden rectangle construction:

Construction is as in my preceding post, with the addition of finding the other intersection of circle ED with line AB (point H), and a perpendicular at H to find point I on line CD.

Geometric Proof that the construction produces golden rectangles:

• HACI and BFGD are congruent by the symmetry of construction.
• Likewise, HBDI and AFGC are congruent by the symmetry of construction.
• It is clear that these pairs of rectangles differ by a square (ABDC); It remains to be shown that these two pairs of rectangles are similar.
• Draw segments HC and CF.
• Because HCF is an inscribed angle on diameter HF of circle E, HCF is a right angle (Thale’s Theorem)
• Angles HAC and CAF are right angles because of the initial construction of square ABCD.
• As triangles AFC and HFC share the angle at F and both have a right angle, they are similar to each other.
• As triangles AHC and HFC share the angle at H and both have a right angle, they are similar to each other.
• Because triangles AFC and AHC are both similar to triangle HFC, they are similar to each other.
• Because of this, rectangles AFGC and AHIC are similar to each other.
• GACI, BHJD, GBDI, and AHJC are all golden rectangles.

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Here's the alternate proof of the original quiz which does not use the Pythagorean Theorem, though it does use the geometric mean theorem. The geometric mean theorem is essentially proven in steps 5 through 8 of the Golden proof above, adding that because AH/AC = AF/AH, AH * AH = AF * AC, or AH is the geometric mean of AF and AC.

• Bisect angle BAC. Note that because AB and AC are both tangent to both circles, this line passes through their centers, designated E and H.
• Identify points D, F, G, and I being the intersections of perpendiculars of AB and AC through E and H, and therefore the points of tangency of those circles with the lines.
• Note that ADE and AFD are congruent, AGH and AIH are congruent, and these pairs of triangles are similar by virtue of having two angles (at A and a right angle) equal.
• Identify point J, being the intersection of AH with the circles. Construct a perpendicular to AH at J, extending past AB and AC. The intersections are points K and L. Because AJK and AJL are congruent (ASA), KJ = JL.
• Note that line KL is tangent to both circles, as are DG and FI. Therefore, DK and KG are both equal to KJ and FL and LI are equal to JL, so all are equal to each other, meaning that K bisects DG and L bisects FI.
• Designate the radius of the small circle as r.
• By definition of k, the large circle has radius rk. EH therefore has length r(k+1)
• Bisect segment EH to identify point M. Note that EM and MH are both length r(k+1)/2.
• Note that AMK and AML are similar to ADE and AHG so that length KM and ML are the average of DE and GH, or r(k+1)/2.
• Draw the circle centered on M passing through E (and therefore H, K, and L).
• Angle EKH is inscribed on circle M and spans diameter EH, so EKH is a right angle.
• By the geometric mean theorem, KJ is the geometric mean of EJ and JH. As EJ is r and JH is rk, KJ is r√k, as are JL, DK, KG, IL, and LF. KL is 2 r√k.
• Construct a perpendicular to AB through L. Designate the intersection with AB as N.
• Because triangles ADE and AGH are similar triangles:
• AD/r = (AD+2r√k)/rk, or AD k = AD + 2r√k, or AD = 2r√k/(k-1), and
• AE/r = (AE + r + rk)/rk, or AE k = AE + r + rk, or AE = r(k+1)/(k-1).
• Because NL is perpendicular to AD, and KL is perpendicular to AE, angle NLK equals angle DAE. Because angle KNL and ADE are both right angles, triangles DAE and NLK are similar. This means:
• NL = KL * AD / AE = 2r√k * 2r√k/(k-1) / (r(k+1)/(k-1)) = 4 r k / (k+1)
• AL = AK = AD + DK = 2r√k/(k-1) + r√k = r√k(k+1)/(k-1)
• As triangles ABC and ANL share the angle at A and have right angles at N and B, they are similar.
• So AC/BC = AL/NL = (r√k(k+1)/(k-1)) / (4 r k / (k+1)) = (k+1)2 /4 √k (k-1).