Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
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Thursday, May 19, 2016
Geometry Problem 1215: Circle, Diameter, Chord, Radius, Midpoint, Measurement
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join A to C, draw DH altitude of ADC => DH meet AB at O
ReplyDeleteHO midleline of ABC => OH=a/2. From tr AOH AH²=x²-a²/4 (1)
From tr ADH AH²=d²-(a/2+x)² (2)
from (1) & (2) x²-a²/4=d²-(a/2+x)² => the conlusion
Let CB and AD extended meet at E. Since AD = CD, BD bisects < ABE, hence DE
ReplyDelete= d and BE = 2x
So 2d^2 = 2x(2x+a) from which
x^2 + ax/2 + a^2/16 = d^2/2 + a^2/16 by completing the square
So (x+a/4)^2 = (a^2+8d^2)/16 and therefore
x = (1/4){sqrt(a^2+8d^2) – a)}
Sumith Peiris
Moratuwa
Sri Lanka
By Ptolemy's theorem, AB·CD+BC·AD=AC·BD
ReplyDelete√((2x)^2-a^2)√((2x)^2d^2)+ad=2dx
√(((2x)^2-a^2)((2x)^2-d^2))=d(2x-a)
((2x)^2-a^2)((2x)^2-d^2)-d^2(2x-a)^2=0
16x^4-8d^2x^2-4 a^2x^2+4 a d^2x=0
4x(2x-a)(2x^2+ax-d^2)=0
x=0 (rejected, ∵x>0) or a/2(rejected, ∵ΔABC is a right triangle∴ AB>BC,ie. 2x>a or x>a/2) or 1/4 (sqrt(a^2+8 d^2)-a) or 1/4 (-sqrt(a^2+8 d^2)-a) (rejected, ∵x>0)
Drop a line from to point D perpendicular onto the diameter AB. The line meets the diameter in the point H.
ReplyDeleteThen by the broken chord theorem AH = (2x+a)/2
The triangles AHD and ADB are similar and we can get :
AH/AD=AD/AB
(2x+a)/2d=d/2x
x^2+ax/2=d^2/2
Solving this quadratic equation we get :
x=-a/4+sqrt(8d^2+a^2)/4