A Very Nice Geometry Challenge | Math Olympiad | 2 Different Methods to Solve

[@holyshit922]

Sine rule in ADC 3/sin(theta) = DC/sin(2theta) Cosine rule in ADC 3^2=(DC)^2+5^2-2DC*5*cos(theta) Cosine rule in ABC x^2=(2DC)^2+5^2-2(2*DC)*5*cos(theta)

[@soli9mana-soli4953]

Thank you for the sin(3α) formula I didn't know. My methos is very similar to your second one, the only difference is that once I found with sines law that DC = 6*cos α I applyed the cosine law on ADC doing: (6 cos α)² = 3² + 5² - 2*3*5*(2cos²α - 1) (being cos 2α = 2cos²α - 1) finding cos α = √ 6/3 and then cos 2α = 1/3 with those values you can find DC and X with cosine law

[@femalesworld2]

3 metod. BD = DC, AD bisector. BAD = DAC = 2© (phita)-???

[@howardaltman7212]

You did this in 2 different ways; I did it 2 other ways. I love these questions that are so rich!

[@User-jr7vf]

I did it like in method 1 up to the similar triangles, then I used a different route to get x.

[@mauriciosahady]

Triangles ABD and ADC have same area. After finding sin theta we can calculate x

[@vcvartak7111]

In 1st method after getting a=2√6 instead of applying Stewart's theorem why not apply Appollonius theorem which is more common since median is known

[@RAG981]

Why do you insist on always quoting the cosine rule in the cos theta form? Yet again it would be simpler here to quote it as a^2=b^2+c^2-2bccosA.

[@olivier9125]

what is b ? 2:30

[@ivannaumov9337]

The correct answer is x=7.

[@josephsalinas6725]

Esse eu fiz !