M is the midpoint of the median AD of the triangle ABC. The ray BM meets AC at N. Show that AB is tangent to the circumcircle of NBC iff BM/BN = (BC/BN)2.
Solution
Applying Menelaus to the triangle ADC, we have (AM/MD)(BD/DC)(CN/NA) = 1, so (CN/NA) = 2. Hence AN/AC = 1/3. Applying Menelaus to the triangle BNC, we have (BM/MN)(AN/AC)(CD/DB) = 1, so BM/MN = 3. That is true irrespective of whether AB is tangent to the circle NBC.
If AB is tangent, then AB2 = AN.AC = 1/3 AC2. Also angle ABN = angle BCN, so triangles ANB and ABC are similar. Hence BC/BN = AC/AB. Hence (BC/BN)2 = 3 = BM/BN.
Conversely, if (BC/BN)2 = BM/BN, then (BC/BN)2 = 3.
Now applying the cosine formula to AMN and AMB and using cos AMN + cos AMB = 0, we have (3AN2 - 3AM2 - 3MN2) + (AB2 - AM2 - BM2) = 0, so AB2 + AC2/3 = AD2 + 3/4 BN2. Similarly from triangles ADC and ADB we get AB2 + AC2 = 2AD2 + BC2/2. So using BN2 = BC2/3 we get 2AB2 + 2/3 AC2 = AB2 + AC2 and hence (AC/AB)2 = 3 = (BC/BN)2. So AC/AB = BC/BN. Note that is not enough to conclude that triangles ABC and BNC are similar, because the common angle C is not between AC and AB. However, we have AN/AB = (1/3) AC/AB = AB/AC, so AB2 = AN.AC, so AB is tangent to the circle NBC.
© John Scholes
jscholes@kalva.demon.co.uk
22 Oct 2002