A geometry problem by Ricky Huang

Geometry Level 3

We have an equilateral $\triangle ABC$ and a random point $O$ in it such that $\angle AOC = x$ and $\angle BOC = y$ . Using lines $AO$ , $BO$ , and $CO$ to make an triangle, which of the answer options cannot be an internal angle of the triangle?

y - 60^\circ

300^\circ - y - x

y - 50^\circ

x - 60^\circ

2 solutions

David Vreken
Jul 1, 2020

Cut out equilateral triangles $\triangle OPP'$ , $\triangle OQQ'$ , and $\triangle ORR'$ , as shown below, and rearrange the remaining triangles as follows, so that $\triangle TUV$ is the new triangle made from $AO$ , $BO$ , and $CO$ :

Then

$\angle T = C'O'A' = \angle AOC - 60° = x - 60°$

$\angle U = C'O'B' = \angle BOC - 60° = y - 60°$

$\angle V = 180° - \angle T - \angle U = 180° - (x - 60°) - (y - 60°) = 300° - y - x$ .

Therefore, the internal angle cannot be $\boxed{y - 50°}$ .

Jeff Giff
Jun 30, 2020

Use logic. The internal angles of a triangle add up to $180^\circ$ , so that we can eliminate option 3 :D
P.S. Do NOT use this approach for problems, since you won’t learn much :(
MATHEMATICAL SOLUTION AWAITS
Hint: rotate AOB about A until AB touches AC. Call the new figure A’O’B’. Study the interior angles of A(A’)OC(B’)O’.

You can rotate AOB about A until AB touches AC. And you can get it.

Isaac YIU Math Studio - 11 months, 2 weeks ago

Oic. Thank you!

Jeff Giff - 11 months, 2 weeks ago

A geometry problem by Ricky Huang

2 solutions

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