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Equal Tangents (Posted on 2010-10-03) Difficulty: 3 of 5
C(UV) denotes the circle with diameter UV.

T(P,QR) denotes the tangential distance |PS|,
where point P lies outside C(QR), point S
lies on C(QR), and PS is tangent to C(QR).

Let A, B, C, and D be distinct, collinear
points in that order.

Construct a point E on line AD such that
|EF| = T(E,AB) = T(E,CD) = |EG|

  Submitted by Bractals    
Rating: 3.0000 (1 votes)
Solution: (Hide)

Construct circles C(AB) and C(CD). Draw a circle
K1 intersecting circle C(AB) in points P1 and Q1 and
circle C(CD) in points R1 and S1. Let T1 be the
intersection of lines P1Q1 and R1S1. Construct the
point T2 similarly. The intersection of lines T1T2
and AD is the desired point E.


When I say construct a circle, I mean that only one
circle is possible from the information given; whereas,
when I say draw a circle, I mean that an infinite number
of circles are possible from the infomation given.


T(X,YZ) is the square root of the "power" of point X
with respect to circle C(YZ).

The locus of points E where T(E,AB) = T(E,CD) is the
"radical line" of the circles C(AB) and C(CD).

The radical line of circles K1 and C(AB) is the line P1Q1.
The radical line of circles K1 and C(CD) is the line R1S1.
The point T1 lies on the radical line of circles C(AB) and
C(CD) (see "radical center"). Similarly, T2 lies on the
radical line of circles C(AB) and C(CD). Therefore, T1T2
is the radical line of circles C(AB) and C(CD).

Thus, E is the desired point.


See Larry's post for an analytic calculation of the distance
from A to E. This distance is constructible and therefore is
an alternate method for constructing point E.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
re(3): ConstructionCharlie2010-10-05 01:05:46
Solutionre: Algebraic solution leads to geometricLarry2010-10-04 22:34:19
re(2): Constructionbrianjn2010-10-04 21:13:39
re(2): Does this work?broll2010-10-04 12:20:04
re: Does this work?Charlie2010-10-04 12:16:49
re: ConstructionCharlie2010-10-04 12:05:22
QuestionDoes this work?broll2010-10-04 02:53:11
SolutionConstructionbrianjn2010-10-04 02:20:35
SolutionNo SubjectAdy TZIDON2010-10-03 15:40:05
re: Algebraic solutionBractals2010-10-03 14:39:12
Some ThoughtsAlgebraic solutionLarry2010-10-03 14:14:53
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