Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
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Sunday, August 19, 2012
Problem 796: Triangle, Altitude, Measurement of Angles
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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°.
ReplyDeleteMigue.
http://img13.imageshack.us/img13/6435/problem796.png
ReplyDeleteDraw 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
Angle HDC = 2x+9
ReplyDeleteBy 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.
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.
ReplyDeletehttp://www.mathematica.gr/forum/viewtopic.php?f=20&t=32532&p=150623
ReplyDeleteIf AD extended meets BC at G, AHGE is concyclic so ADGC is also concyclic
ReplyDelete< AHE = 90 = (x+9) + (2x+6)
Therefore x = 25
Sumith Peiris
Moratuwa
Sri Lanka