CEMC Fermat, 2012

Geometry Difficulty 3.4 AMC 10/12 Find the answer Canada

In the diagram, $\triangle QUR$ and $\triangle SUR$ are equilateral triangles. Also, $\triangle QUP$ , $\triangle PUT$ and $\triangle TUS$ are isosceles triangles with $PU=QU=SU=TU$ and $QP=PT=TS$ . The measure of $\angle UST$ , in degrees, is

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Since $\triangle QUR$ and $\triangle SUR$ are equilateral, then $\angle QUR = \angle SUR = 60^\circ$ .

Since $QU=PU=TU=SU$ and $QP = PT=TS$ , then $\triangle QUP$ , $\triangle PUT$ and $\triangle TUS$ are congruent.

Thus, $\angle QUP = \angle PUT = \angle TUS$ .

The angles around point $U$ add to $360^\circ$ .

Thus, $\angle SUR + \angle QUR + \angle QUP + \angle PUT + \angle TUS = 360^\circ$ and so $60^\circ + 60^\circ + 3\angle TUS = 360^\circ$ or $3\angle TUS = 240^\circ$ or $\angle TUS = 80^\circ$ .

Since $\triangle TUS$ is isosceles with $TU = SU$ , then $\angle UST = \angle UTS$ .

Since the angles in $\triangle TUS$ add to $180^\circ$ , then $\angle TUS + \angle UST + \angle UTS = 180^\circ$ .

Therefore, $80^\circ + 2\angle UST = 180^\circ$ and so $2\angle UST = 100^\circ$ or $\angle UST = 50^\circ$ .

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