Saturday, October 27, 2012

Problem 819: Quadrilateral, Triangle, Angles, 30 degrees

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 819.

Online Geometry Problem 819: Quadrilateral, Triangle, Angles, 30 degrees

21 comments:

  1. Triangle ABC is isosceles, AC = 2ABcos40
    Using sine rule on triangle ACD,
    AC/sinx = CD/sin30
    2ABcos40/sinx = CD/sin30
    cos40 = sinx
    x = 50

    ReplyDelete
    Replies
    1. why x not= 130?
      I believe x=50 or 130.

      Delete
  2. In triangle ABC:
    AC = 2 AB cos(40) ---- (1)
    In triangle ACD:
    AC = CD sin(x)/sin(30) = 2 CD sin(x) ---- (2)
    AB = CD;
    Then by (1) and (2):
    sin(x) = cos(40) = sin(50)
    x = 50

    ReplyDelete
    Replies
    1. Why x=50 only?
      I believe x=50 or 130.

      Delete
  3. To Anonymous: according to the figure angle x is acute, therefore x=50. If D' is on AD so that AB=BC=CD', then angle x is obtuse (x=130).

    ReplyDelete
    Replies
    1. Antonio, did you judge the answer according to figure?

      Delete
    2. To Anonymous: I think the problem has two solutions. First, according to the figure, when the angle x is acute and the second, when the angle x is obtuse.
      Thanks.

      Delete
  4. Problem 819: Try to use elementary geometry (Euclid's Elements).

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  5. Let E be the circumcenter of ΔACD.

    Then
    ∠CED = 2×∠CAD = 60°, CD = CE
    ⇒ ΔCDE is equilateral
    ⇒ AB = BC = CD = DE = CE = AE
    ⇒ ABCE is a rhombus
    ⇒ ∠AEC = ∠ABC = 100°
    ⇒ ∠ADC = 1/2×∠AEC = 50°

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  6. draw the altitudes BM of triangle ABC,CH of triangle ACD
    CM=AC/2 (ABC is isoscele)
    CH=AC/2 (triangle 30-60-90)
    the right triangles BMC and CHD are congruent
    x=50

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  7. Prof Radu Ion,Sc.Gim.Bozioru,BuzauOctober 29, 2012 at 10:08 AM

    Costruind simetricul punctului C fata de dreapta AD si notandu-lcu P obtinem :
    ΔACP echilateral si ΔACP=ΔPCD(LLL);∠CDP = ∠ABC = 100°,AD mediatoarea seg AD =>AD bisectoarea ∠CDP => ∠CDA = 50°
    Prof Radu Ion,Sc.Gim.Bozioru,Buzau

    ReplyDelete
  8. Draw BE perpendicular to AC and CF perpendicular to AD. Join EF.
    BE bisects AC, since AB = BC
    E being the midpoint of the hypotenuse AC of the right triangle AFC,
    EA = EC = EF and consequently ΔEFC is equilateral.
    Now consider the right triangles BEC and DFC.
    By Pythagoras, DF^2 = CD^2 - CF^2 = BC^2 - CE^2 = BE*2.
    So DF = BE and consequently the right triangles CDF and CBE are congruent.
    Follows x = ∠FBC.
    Note that ΔABC is isosceles (with base angles FCB, FAB each 40°),
    Hence x = ∠FBC = 90° - ∠FCB = 90° - 40° = 50°


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  9. x=50deg if AD>AC; or x=130deg IF AD<AC.

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  10. Let BA=BC=CD=a and AC=d
    Using sine rule,
    In triangle ABC, a/sin 40=d/sin 100…(1)
    In triangle ACD, a/sin 30=d/sin x…(2)
    Using (1),(2), we get, sin 100/sin40=sin x/sin 30
    sin 80/sin40=sin x/sin 30
    2(sin 40)(cos 40)/sin40=sin x/(1/2)
    cos 40=sin x
    sin (90-40)=sin x
    sin 50=sin x
    50=x.

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  11. Let E be the point on AD such that AB= BE.
    ang. AEB= 700 (∆ ABE is isosceles)
    ang ABC= 1000= ang. ABE+ ang. EBC → ang EBC= 600
    Since BE=BC (by our construction) so,
    ang. BEC= ang. ECB = 1800-600/2= 600
    This implies ∆ BEC is equilateral. → CD= CE.
    so ang. X= ang. CED = 1800-(700+600) = 500

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  12. Sorry for the extra zeros in the solution, they all refers degrees.
    Thank You.

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  13. En la figura adjunta resuelvo el problema sin usar CD...:D
    http://www.subirimagenes.com/otros-ahorasi-8277920.html

    By Tony García.

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  14. http://www.mathematica.gr/forum/viewtopic.php?f=20&t=32307&p=149499

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  15. Draw altitudes BH and CG. Then by 30-60-90 Triangle and congruence x=50

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  16. Another proof. Find E on AD such that AB = BE. Then < EBC = 60 and Tr. EBC is equilateral, Tr. ECD is isoceles and x = 50.

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  17. Let D' on AD so that CD'=CD and <AD'C is obtuse and let B' be the circumcenter of tr. AD'C. Clearly tr CB'D' is equilateral and CB'=AB'=CD'=CB=AB, so B'=B, thus <AD'C=180-m(<ABC)/2=130 degs. and, as constructed, <ADC=<CD'D=180-m(<AD'C)=50 degs.

    Best regards

    ReplyDelete