Sunday, August 19, 2012

Problem 796: Triangle, Altitude, Measurement of Angles

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

Click the figure below to see the complete problem 796.

Online Geometry Problem 796: Triangle, Altitude, Measurement of Angles.

5 comments:

  1. Al prolongar AD intersecta a BC en ángulo recto por ser el ángulo C=90-x. Así D es el ortocentro del triángulo y debe cumplirse que: x+6+x+x+9=90 y por tanto x=25°.

    Migue.

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  2. http://img13.imageshack.us/img13/6435/problem796.png

    Draw lines per attached sketch
    Quadrilateral AHEB is cyclic => AE ⊥ BC
    Quadrilateral ACEF is cyclic => CF ⊥ AB
    In AHEB angle BAE= ½(180-4x-12)=84-2x …….(1)
    In ACEF angle BAE=x+9…… (2)
    From (1) and (2) we have 84-2x=x+9 => x=25

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  3. Angle HDC = 2x+9

    By comparing triangle ADH and DHC, we have HC/AH = tan(2x+9)*tan(x)
    By comparing traingel ABH and HBC, we have HC/AH = tan(x) / tan (x+6)

    Therefore, tan(2x+9)*tan(x) = tan(x) / tan (x+6) => tan(2x+9)*tan(x+6) = 1

    Using product to sum, of sine and consine, we have
    cos((2x+9)-(x+6)) - cos((2x+9)+(x+6)) = cos((2x+9)-(x+6)) + cos((2x+9)+(x+6))
    cos((2x+9)+(x+6))=0
    3x + 15 = 90
    x = 25.

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  4. Extend AD to meet BC at E. Each of the angles HAE, HBE being x, the 4 points H, A, B, E are con-cyclic. It then follows that angle BDA = BHA = 90. So D is the ortho-center of triangle ABC and CD is perpendicular to AB. Hence the sum of the angles BAE and ABE is 90 degrees, which means x + 9 + x + x + 6 = 90, 3x = 75, x = 25 degrees.

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  5. http://www.mathematica.gr/forum/viewtopic.php?f=20&t=32532&p=150623

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