Friday, September 16, 2016

Geometry Problem 1261: Cyclic Quadrilateral, Circle, Diameter, Midpoint, Measurement

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

This entry contributed by Sumith Peiris, Moratuwa, Sri Lanka.

Click on the figure below to view more details of problem 1261.


Geometry Problem 1261: Cyclic Quadrilateral, Circle, Diameter, Midpoint, Measurement

11 comments:

  1. https://goo.gl/photos/tDgKutmrL4Bn9Lur8

    Draw points O, E, F, G per sketch
    Note that OE ⊥AB , AF⊥ EF and triangles EBC, AFE are 45-45- 90 triangles
    We also have AF/FC= CD/AD= 1/3 => EC//AE
    And BE/BA=BC/BE= 1/2

    ReplyDelete
    Replies
    1. Dear Peter - How is AF/FC = CD/AD?

      Delete
    2. Since AB=2BC=> BC=BE=AE ( see sketch)
      Triangles AFE, EGB and BGC are 45-45-90 triangles and are congruent ( case ASA)
      so FA=FE=EG=GC => AF/FC= 1/3

      Delete
  2. Palai Renato
    We put AB=1, BC = 2 AC= √5,
    We put CD=x we have x+(3x)2=√5, and therefore x= 1/√2
    <BAC =45 (regle of sum of tangent), CD=DE and CE=1
    CE=CB and C is midpoint of BE
    Reply

    ReplyDelete
  3. ABCD is cyclic, with AC being diameter. Hence, Angle ABC and Angle ADC are right angles, by Thales' theorem.

    By pythagoras’ theorem, it’s evident that BC = AC/sqrt(5), AB = 2AC/sqrt(5), CD = AC/sqrt(10), AD = 3AC/sqrt(10). We need the sine and cosines of angle BAC, angle CAD.
    If we let O be the circumcenter of ABCD, then angle BOC = 2 angle BAC, angle COD = 2 angle CAD. This is obvious high school geometry.

    From here, we use complex numbers.
    Let the circle on ABCD be the unit circle and c = 1, a = -1.
    With a bit of trig (especially double angle formula), b = cis(angle BOC) = cis(2 angle BAC) = (3/5) + i(4/5). Similarly, d = cis(-angle COD) = (4/5) – i(3/5).

    Now let’s make a point F such that C is the midpoint of B and F. This indicates (f + b)/2 = c, which rearranges to f = 2c – b. We compute f = (4/5) – i(4/5).

    We now assert that A, D and F are collinear. This is proved by checking that (a – f)/(a – d) is a real number (justification is De Moivre theorem; (a - d) multiplied by a real number would make (a-f) iff a-f and a-d make an angle of 0 or 180 degrees with each other).

    (a – f)/(a – d) = (12/5 – i(4/5))/((-9/5) – i(3/5)) = (4/5) / (-3/5) = -4/3 which is definitely a real number. So we conclude that F is on line AD and BC (as C is midpoint of BF).

    But we know that E is the intersection of AD and BC. This means E = F, and the proof is complete.

    ReplyDelete
  4. Problem 1261
    Is <BAC+<CAD=45, so <CED=45 then CD=DE. Let the points K, L on the side AD such that
    AK=KL=LD=AD/3=DC=DE. So AL=LE and BL is perpendicular at AE so BL//CD. Therefore
    BC=CE (LD=DE).
    APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

    ReplyDelete
  5. Palai Renato,
    Reply to Sumith Peiris to request to motivate why angle.. is 45°.
    If we calla <BAC= and <CAD= , <BAD= + , bat
    tang ( + )=(tang + tang )/(1 - tang tang ) and introducing the values of and , we have: (2+3)//1-2x3)= -1, tang 1 = 45°
    Reply

    ReplyDelete
  6. Let CD = 1
    => AD = 3
    => AC^2 = CD^2 + AD^2 = Sqrt(10)

    Similarly solving for AB & AC in the right triangle ABC,
    we get BC = sqrt(2) --------(1)
    and AB = 2*Sqrt(2)

    Let CE = x and DE = y
    As CE^2 = DE^2 + CD^2
    => 1+y^2 = x^2 ----------- (2)

    As ED.EA = EC.EB (From Secant-Secant Rule)
    => y.(3+y) = x.(x+Sqrt(2))
    => y^2+3y = x^2+Sqrt(2).x
    => 3y = (x^2-y^2) + Sqrt(2).x
    => 3y = 1 + Sqrt(2).x (from (1) )
    => y = [1+Sqrt(2).x]/3 ----------(3)

    Substituting the value of y in Eq(2)

    => 9 + (1+2x+2Sqrt(2)x) = 9x^2
    => 7x^2 - 2Sqrt(2)x -10 = 0

    Therefore x = (2Sqrt(2) + Sqrt(8+288))/14 = 14.Sqrt(2)/14 = Sqrt(2) = BC (from (1) )

    Hence C is midpoint of BE

    ReplyDelete