Q160 | Math Olympiad | Geometry | Circle | Power of a Point

[@MarieAnne.]

Nice solution. For those unfamiliar with Power of a Point theorem, we can use Pythagorean theorem by constructing right triangles. Drop perpendicular from A to BC at point P. Since △ABC is isosceles (with AB = AC), altitude AP bisects BC. Let AP = h, BP = CP = a, CD = b. Then BD = 2a + b Using Pythagorean theorem in △APC, we get: h² + a² = 2² (1) Using Pythagorean theorem in △APD, we get: h² + (a+b)² = 4² (2) Subtracting (1) from (2) we get: (a+b)² − a² = 16 − 4 a² + 2ab + b² - a² = 12 2ab + b² = 12 (2a+b) * b = 12 BD * CD = 12

[@singher5204]

You could also get the product by using Stewart's Formula to find the Cevian of a Triangle.

[@zdrastvutye]

there is a symmetry and the result does not depend on "h=" in line 20: 10 print "reach the stars-math olympiad-geometry":dim x(1,2),y(1,2) 20 l1=2:l2=4:h=1.5:print "h="h:lh=sqr(l1^2-h^2):l3=sqr(l2^2-h^2)-lh 30 print "Das ergebnis BD*CD="; (2*lh+l3)*l3 40 x(0,0)=0:y(0,0)=0:x(0,1)=2*lh:y(0,1)=0:x(0,2)=lh:y(0,2)=h 50 x(1,0)=2*lh:y(1,0)=0:x(1,1)=2*lh+l3:y(1,1)=0:x(1,2)=lh:y(1,2)=h 60 masy=850/h:masx=1200/(2*lh+l3):if masx run in bbc basic sdl and hit ctrl tab to copy from the results window.