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I was recently asked to help with the following geometry problem:
Given a circle centered at point A with radius AB and a point C outside circle AB, using a compass and unmarked straightedge, find the lines tangent to circle AB that intersect point C.
Figure 1: Tangents of x^2+y^2=4 intersecting (8,3)
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I decided to try to solve it without using any outside references--just my own understanding of geometry. My geometry curriculum in school, while covering some theorems and "pure" constructions, was largely analytical in nature. As a result, I was able to use analytical geometry to determine fairly quickly where the tangent points on the circle would lie, but I found it quite difficult to produce a construction without using angle measurements. It took a few days for me to iron out a solution.
Finally, using similar triangles, I devised the following construction.
We begin by drawing line AC, which intersects circle AB at two points D and E. Draw circle DE (centered at point D with radius DE) and circle ED. These two circles intersect at two points: F and G. Draw line FG. This is a standard construction that creates a line perpendicular to line AC that passes through point A.
Figure 2: Line FG is perpendicular to line AC and intersects A
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Line FG intersects circle AB at two points. Without loss of generality, call one of these points H. Draw line CH; this intersects circle AB at point I. Draw line AI, which intersects circle AB at point J. Draw line HJ; this intersects line AC at point K.
This part of the construction creates two right triangles: ∆ACH and ∆HCK. Note that these triangles are similar.
Figure 3: ∆ACH and ∆HCK are similar
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Draw circle HK; this intersects lineAC at point L. Draw circle LA, which intersects line AC at point M. Draw circle MA and circle AM. These two circles intersect at points N and O. Draw line NO; this intersects circle AB at points P and Q. Line CP and line CQ are then the tangents to circle AB that intersect C.
Figure 4: Line CP and line CQ are tangents
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The key to this construction is using similar triangles to construct the point L. To see why this works, we'll work backwards from the tangents.
Fix the center of the circle AB at the origin, set its radius to 1, and suppose point C is located at (x,0). Then ∆ACP is a right triangle whose short leg is AP=1 and whose hypotenuse is AC=x. Segment LP is an altitude on the hypotenuse of ∆ACP, so ∆LPA is similar to ∆ACP. The hypotenuse of ∆LPA is AP=1, which means the length of the short leg AL is 1/x. We now have a new problem to solve: construct the length 1/x, from which we can construct the tangents.
Enter the right triangle ∆ACH. This triangle is not similar to ∆ACP: its short leg is AH=1 and its long leg is AC=x, so its hypotenuse is CH=sqrt(x^2+1). It is, however, similar to ∆HCK, since line HK is constructed to be perpendicular to line CH. The long leg of ∆HCK is CH=sqrt(x^2+1), so the length of the hypotenuse CK is:
x/sqrt(x^2+1)=sqrt(x^2+1)/CK
xCK=x^2+1
CK=x+1/x
Because the length of AC is x, the length of AK is 1/x, our desired length.
The only steps left are to mirror point K on the other side of line AH, giving us L, and to construct a perpendicular line to line AC that intersects L. The full construction takes 12 steps to find points P and Q.
Figure 5: The full construction has many steps
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After I finished my construction, I looked up the problem online to see if there was a "standard" construction--and it didn't take long to find one. The following construction takes only five steps.
Begin by drawing line AC . Draw circle AC and circle CA; these intersect at points D and E. Draw line DE, which intersects line AC at point F. Draw circle FA; this intersects circle AB at points G and H. Line CG and line CH are the tangents to circle AB that intersect C.
Figure 6: The standard construction takes five steps
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Unlike my construction, the standard construction uses purely geometric concepts. Specifically, point F is the midpoint of AC, constructed using the perpendicular bisector of AC. As noted, A, C and the tangent point G form a right triangle, of which AC is the hypotenuse. An important property of a right triangles is that the center of its circumcircle is located on the midpoint of its hypotenuse; the standard construction uses this fact to determine the location of G (and H) using F directly.
To me, this is a good example of different methods of problem solving. A weakness of mine is that I tend to take a "one tool fixes all" approach to problem solving: if one method works, even if it's inefficient, well, at least it works. In this case, I'm well-versed in analytical geometry, and I was able to use it to find a solution, clunky as it is. Were I better versed in pure geometry and the properties of polygons--a seemingly-unrelated concept--I might have been able to find this construction on my own.
Many real-world problems require optimization and efficiency, sometimes even at the cost of correctness in corner cases. In order to find such optimizations, it's often necessary to attack a problem from many different angles. This is the very definition of thinking outside the box.
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[Last updated: 2022-09-07]
