Symmetrical Spiky Polygon

Geometry Level 1

The figure to the right consists of 6 straight lines. Suppose the following is known:

$|\overline{A_1B_1}| = |\overline{A_2B_2}|,\ |\overline{A_1B_2}| = |\overline{A_2B_1}|,\$ and $|\overline{B_1C}| = |\overline{B_2C}|.$
$\overline{A_1B_1} \perp \overline{A_2B_1}$ and $\overline{A_1B_2} \perp \overline{A_2B_2}.$
The three acute angles $A_1, A_2,$ and $C$ have congruent measures.

What is the measure of $\angle CB_1A_2?$

30^{\circ}

45^{\circ}

60^{\circ}

None of the measurements It depends on the setup of the points

2 solutions

Marta Reece
Dec 29, 2017

The three angles being the same requires all five points to be on a circle.

The two right angles mean that points $A_1$ and $A_2$ form a diameter of the circle with point $C$ half way between them.

If $O$ is the center of the circle, the angle $\angle AOC=90^\circ$ .

So the angle $\angle CB_1A_2=\frac12 \angle AOC=\boxed{45^\circ}$

Nice and fast! Thanks for the solution!

Michael Huang - 3 years, 5 months ago

John Miller
Jan 6, 2018

The distance between B1 and B2 is irrelevant, thus those points could merge into a single B12 point directly below C. Such a merge would then cause the four AB lengths (A1B1, A2B1, A1B2, and A2B2) to become only two lengths, A1B12 and A2B12. By definition, those lengths are the same. Angles A1, A2, and C become 0 degrees. The CB lengths, CB1 and CB2 become a single line CB12, which bisects angle A1B12A2 resulting in angle CB12A2 being 45 degrees.

Symmetrical Spiky Polygon

2 solutions

0 pending reports