Can you do this without trigonometry?

Geometry Level 4

A circle, centre O O , has A B AB as a diameter. Let C C be a point on the circle different from A A and B B , D D be the point on A B AB such that ∠ C D B = 9 0 ∘ \angle CDB = 90^{\circ} and M M be the point on B C BC such that ∠ B M O = 9 0 ∘ \angle BMO = 90^{\circ} . If D B DB is 3 × O M 3 \times OM , calculate ∠ A B C \angle ABC in degrees.


The answer is 30.

Sharky Kesa by Sharky Kesa , 19, United Kingdom

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1 solution

Let OM=n, DB= 3n, BO=OC=r, BM=MC=x, ∴ D O = 3 n − r . U s i n g P y t h a g o r a s , Δ s C D O , a n d C D B , r 2 − ( 3 n − r ) 2 = C D 2 = ( 2 x ) 2 − ( 3 n − r ) 2 . B u t x 2 = r 2 − n 2 . ⟹ r 2 − ( 3 n − r ) 2 = 4 ( r 2 − n 2 ) − ( 3 n − r ) 2 . Solving the quadratic for r, and noting that r>0, take + sign. r = 2 n . ⟹ 3 0 o = S i n − 1 1 2 = S i n − 1 n r = ∠ A B C . \text{Let OM=n, DB= 3n, BO=OC=r, BM=MC=x, } \therefore~DO=3n-r .\\ Using ~~Pythagoras,\\ \Delta s ~CDO,~~and~~CDB,~~r^2 - (3n-r)^2=CD^2=(2x)^2 - (3n - r)^2.\\ But~x^2=r^2 - n^2.\\ \implies~r^2 - (3n-r)^2=4(r^2 - n^2) - (3n - r)^2.\\ \text{Solving the quadratic for r, and noting that r>0, take + sign.} \\ r=2n.~~\implies~~{\huge \color{#D61F06}{30^o}}=Sin^{-1}\dfrac 12=Sin^{-1}\dfrac nr=\angle ABC.

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