The following problem is certainly appropriate for later in the year when geometry students reach this topic but it can also be used to review a considerable number of essential ideas in preparation for SATs, ACTs or just review in general. It's at the top end on the difficulty scale for these tests, but it's far from the AMC Contest!
Clarifications: Figures are not drawn to scale and the measure of ∠TAU is given in each diagram.
For each of the figures above, determine the following:
(a) the radius of each circle
(b) the length of minor arc TU in each circle
Have fun discovering a variety of approaches!
Variations? Generalizations? Choosing an angle other than special cases like 60 or 90 generally requires trig -- not that there's anything wrong with that!
Monday, September 15, 2008
Reviewing Geometry for Class or SATs - Just a little tangent exercise?
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2 comments:
The simplest method is based the the fact that tangents are perpendicular to radii. Consider Figure 1.
Let O be the center.
∠ OTA = ∠OUA = 90°.
∠TOU = 120°, since OTAU is a convex quadrilateral and its angles sum to 360°.
∠OTU = ∠OUT since TOU is isosceles.
∠ATU = ∠AUT by complementary angles.
∠ATU = ∠AUT = 60°, since ATU is a triangle.
Also, ∠OTU = ∠OUT = 30°.
ATU is equilateral, so TU = 12.
OTA and OUA are congruent, by SSS.
∠TOA = ∠UOA by corresponding angles, so OA bisects ∠TOU.
∠TOA = 60°, half of ∠TOU.
Let X be the intersection of OA and TU. Then, OTX and OUX are 30°-60°-90° triangles. They are congruent, and X is then the midpoint of TU.
TX = 6, and so OT = 4√3.
Did I miss anything?
Eric, you rarely miss anything!
Some of my students used your approach. A few others did not draw TU, choosing to work directly with triangle ATO, which they concluded was a 30-60-90. From there, the radius follows as well.
The reason I asked for the arc length was to review additional aspects of circle geometry. Overall, not a very difficult problem, but several struggled because their geometry skills were rusty. This is highly typical of what happens when students are away from this course for any length of time.
The 90 degree case is also interesting in that it relates to the issue of inscribing a circle in a square.
As the tan-tan angle increases and approaches 180, it is instructive for students to see how fixing the length of the tangent segment results in increasing the radius. In general, the radius equals (12)(tan (1/2)∠TAU). The curvature of the circle is approaching that of a straight line. The distance between point A and the circle is given by
12 ⋅ cos((1/2)∠TAU) which approaches zero as the half-angle approaches 90. Students need to draw several of these cases to get a feel for the relationships.
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