Department of Mathematics

Winter 2009 MATH 330B/539 Geometry Homework 2 Hints

• due Friday, January 16, 2009
• 4.2#3, 7, 10, 14
• 4.3#1, 5
• no Geogebra project for this homework set

• 4.2#3
  • typo: Must assume AB = BC.
  • Apply Theorem 4.7 to triangle BDA and also to triangle BDC. You will end up with two equations involving ratios. Combine them into one equation. Then use Theorem 4.3.
• 4.2#7
  • Prove that AH = HM. Using this fact, you should be able to prove what you need to prove about AB/AH.
  • Then Apply Theorem 4.7 to triangle BCH. You should be able to recognize the resulting ratio as being that of a famous triangle.
• 4.2#10 Apply Th. 4.12 to segments AB and EB. Then show that triangle(CBA) ~ triangle(DBC). Then use transitivity.
• 4.2#14
  • Prove by contrapositive. That is, prove that if line(AT) is not tangent to the circle, then (AT)^2 does not equal (AB)(BC). You will need to use an early theorem in the book that says that a tangent to a circle touches the circle exactly once. (Find that theorem.) So in other words, if line(AT) is not tangent to the circle, then line(AT) must touch the circle at an additional point besides point T.
  • You will need to use Theorem 4.12.
• 4.3#1
  • Let x1 = length of base of triangle 1, let x2 = length of base of triangle 2, let x3 = length of base of triangle 3. Let S = area(triangle(ABC)).
  • Using Theorem 4.14, get equation involving the ratios S1/S and x1/AC. Similarly for the other three triangles.
  • For the main computation, get an equation expressing the length AC in terms of x1, x2, x3.
  • Divide both sides of this equation by AC. The right side of this equation should now be a bunch of ratios of lengths.
  • Replace these expressions with square roots of ratios of areas.
  • Finally, multiply both sides of this equation by sqrt(S).
• 4.3#5
  • Consider the original hexagon, called hex1. The sides of this hexagon have length a. But this hexagon can be divided up into six equilateral triangles, each of which has three sides of length a. This tells us that in hex1, the distance from the center to a vertex is a.
  • Now consider one of the six triangles that makes up hex1. Find the height h of one of these triangles, in terms of a. Notice that the foot of the altitude will be at the midpoint of the base. This midpoint will be one of the vertices of hex2. In other words, the height h is actually the distance from the center to a vertex of hex2. So the height h is also the length of one of the sides of hex2.
  • From the known lengths of the sides of hex1 and hex2, you should be able to determine the ratio area(hex1)/area(hex2).
  • Imagine repeating this process until you get to hex6.

Suggested problems:

• 4.2#2, 4, 5, 8a, 16
• 4.3 #7

• 4.2#4
  • The first job is to get an equation that expresses the ratio BO/EO in terms of a, b, c.
  • Apply Th 4.7 to triangle(ABE) with bisector ray(AO) Call the result Equation 1.
  • Apply Th 4.7 to triangle(ABC) with bisector ray(BE) Call the result Equation 2.
  • Eliminate EC in Equation 2 by using the known relationship between EC, AC, and b. Call the result Equation 3.
  • Solve Equation 3 for AE, and call the result Equation 4.
  • Substitute Equation 4 into Equation 1 and simplify.
  • Produce analogous expression for the ratio AO/DO in terms of a, b, c.
• 4.2#16 Show that point B lies on line(AC). To do this, suppose that line(AC) intersects the left circle at point B1 and intersects the right circle at point B2. Then show that points B1 and B2 are actually the same point.
  • Apply Corollary 4.1 to left circle to get an equation involving AC, B1C, and CD.
  • Apply Corollary 4.1 to right circle to get an equation involving AC, B2C, and CE.
  • Apply Theorem 2.6 to segments CD and CE.
  • Do algebra to get the equation B1C = B2C. Draw a conclusion from this equation.
• 4.3#7 does not use any techniques of from Section 4.3, but it is a fun problem that is not too hard.
  • Given parallelogram(ABCE), labeled counterclockwise with AB as the base. Let d1 = AC, let d2 = BD, let E be the point on line(AB) such that line(CE) is perpendicular to line(AB), and let F be the point on line(AB) such that line(DF) is perpendicular to line(AB). Let h = CD = DF>. Observe that AF = BE. Call this length x.
  • Apply Pythagorean Theorem to triangle(AEC).
  • Apply Pythagorean Theorem to triangle(BFD).
  • For part (b), it will help to rewrite the equation as (a^2 + b^2) = (d1^2 + d2^2)/2.

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Last updated March 8, 2009.