The points $A,B,$ and $C$ lies on the circumference of the unit circle. Furthermore, it is known that $AB$ is a diameter of the circle and \[\frac{|AC|}{|CB|}=\frac{3}{4}.\] The bisector of $ABC$ intersects the circumference at the point $D$. Determine the length of the $AD$.
Problem
Source: Finland 2013, Problem 3
Tags: trigonometry, geometry unsolved, geometry
sunken rock
03.05.2013 00:12
$\Delta ABC$ is of $3-4-5$ type, so $AC=3k, BC=4k, AB=5k$. Ptolemy for $ABCD$ gives $BD=3AD\implies AD=k\sqrt{2.5}$ ($BD\bot AD$). Best regards, sunken rock
professordad
03.05.2013 01:10
We can also use trig: note that the triangle is 3/5-4/5-1. Let $\theta$ face the 3/5 side of the triangle. Then the answer we're looking for is just $\sin{\left(\tfrac{\theta}{2}\right)} = \sqrt{\tfrac{1 - \cos{\theta}}{2}}$. Note that $\cos{\theta} = \tfrac{4}{5}$ so the answer is $\tfrac{1}{\sqrt{10}}$.
5/6: highly likely: After seeing sunken rock's solution
sayantanchakraborty
12.09.2013 21:57
Let $AC=3k,BC=4k$ Applying Pythagoras theorem for ABC we get $k=\frac{1}{5}$. Apply Ptolemy's theorem and write $BD=(1-AD^2)^{\frac{1}{2}}$. We get $AD=\frac{1}{10}^\frac{1}{2}$ Maths is the doctor of science... Sayantan...
crastybow
06.03.2015 03:48
Hello, isn't $k = \tfrac{2}{5}$? Also, the answer does not seem to be $\tfrac{1}{\sqrt{10}}$. Following sunken rock's solution, we have $AD = \tfrac{\sqrt{10}}{5}$.
MSTang
06.03.2015 04:02
I agree with that - $AB = 2,$ not $1,$ since $AB$ is a diameter of the unit circle.
Kinzoku99
04.08.2017 19:18
crastybow wrote: $AD = \tfrac{\sqrt{10}}{5}$ I agree and we dont even need Ptolemy here... It can be done using Pythagorean theorem and Inscribed angle theorem only if we make a good drawing.