Saturday, July 18, 2015

Geometry Problem 1140: Triangle, Circumcircle, Circle, Center, Collinear Points

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the diagram below to enlarge it.

Online Math: Geometry Problem 1140: Triangle, Circumcircle, Circle, Center, Collinear Points.

Posted by Antonio Gutierrez at 3:28 PM
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4 comments:

  1. Peter TranJuly 31, 2015 at 11:36 AM

    http://s9.postimg.org/lsdftz333/problem_1140.png

    Let x= ∠ (CB2A1) and y= ∠ (BAC)
    Let A1C1 cut BC at G
    Observe that Circle O and circle B intersect at A1C1 => OB is perpendicular bisector of A1C1 and
    ArcBC1= ArcBA1
    1. In circle O we have ∠ (BAC)= ½( Arc BA1+ Arc A1C)
    In circle O we have ∠ (BGC1)=1/2( Arc BC1+Arc A1C)
    But Arc BC1=Arc BA1 => ∠ (BAC)= ∠ (BGC1)=y= ∠ (B1C1C2)

    2. In circle A1CB2 we have
    ∠ (A1B2C)= ∠ (A1A2C)= ∠ (BA2D)=x
    Quadrilateral C1B1B2A1 is cyclic so ∠ (A1C1B1)= ∠ (A1B2C)=x= ∠ (A1A2C)= ∠ (BA2D)

    3. We have ∠ (DC1A1)=y-x= ∠ (DA1C1) => DC1=DA1 => D will be on perpendicular bisector of A1C1
    And B, D and O are collinear

    ReplyDelete
  2. AnonymousAugust 11, 2015 at 2:29 PM

    Beautiful. Any idea on the second part of the problem, which is to prove, B, E, F are colinear?

    ReplyDelete
  3. Ivan BazarovJuly 20, 2016 at 12:22 PM

    <C1A1B1=a и <A1C1B2=c. E'- инверсия E относительно окружность B, которая конечно лежит на EB. Значит что угол <BE'B1=<BB1E=90+a, <BE'B2=<BB2E=90+C и <B1E'B2=360-(90+a)-(90+c)=180-a-c. <AB1C2=(<AB1C1=<C1A1B2)+(<C2B1C1=<C1A1B)=90-c и <CB2A2=(<A1B2C=<B1C1A1)+(<A1B2A2=<BC1A1)=90-a. Поэтому <B1FB2=180-(90-a)-(90-c)=a+c. Точки B1, E', B2, F образуют вписанный четырехугольник, так как <B1E'B2+<B1FB2=180-a-c+a+c=180. Тогда мы знаем что <B1E'F=<B1B2F=90-a. Помните что <B1E'B=90+a, так что <BE'F=<BE'B1+<FE'B1=90+a+90-a=180 и B, E', F лежат на одной прямой. По определению E', E тоже лежит на тот прямой.

    ReplyDelete
  4. Ivan BazarovJuly 20, 2016 at 12:32 PM

    Let <C1A1B1 = a, <A1C1B2 = c, and E' be the inversion of E with respect to the circle B. This means that the angle <BE'B1 = <BB1E = 90 + a, <BE'B2 = <BB2E = 90 + C and <B1E'B2 = 360- (90 + a) - (90 + c) = 180-a-c. <AB1C2 = (<AB1C1 = <C1A1B2) + (<C2B1C1 = <C1A1B) = 90-c, and <CB2A2 = (<A1B2C = <B1C1A1) + (<A1B2A2 = <BC1A1) = 90-a. Therefore <B1FB2 = 180- (90-a) - (90-c) = a + c. The points B1, E ', B2, F form an inscribed quadrilateral as <B1E'B2 + <B1FB2 = 180-a-c + a + c = 180. Then we know that <B1E'F = <B1B2F = 90-a. Remember that <B1E'B = 90 + a, so that <BE'F = <BE'B1 + <FE'B1 = 90 + a + 90-a = 180 and B, E ', F are collinear. By the definition of E ', E also lies on the line.

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