## Thursday, May 19, 2016

### Geometry Problem 1215: Circle, Diameter, Chord, Radius, Midpoint, Measurement

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

Click the figure below to view more details of problem 1215. 1. join A to C, draw DH altitude of ADC => DH meet AB at O
HO midleline of ABC => OH=a/2. From tr AOH AH²=x²-a²/4 (1)
from (1) & (2) x²-a²/4=d²-(a/2+x)² => the conlusion

2. Let CB and AD extended meet at E. Since AD = CD, BD bisects < ABE, hence DE
= 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

3. √(((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)

4. Drop a line from to point D perpendicular onto the diameter AB. The line meets the diameter in the point H.

Then by the broken chord theorem AH = (2x+a)/2

The triangles AHD and ADB are similar and we can get :