2001 CEMC Gauss (Grade 8) Problems/Problem 21

Problem

Lines $PS$ , $QT$ and $RU$ intersect at a common point $O$ , as shown. $P$ is joined to $Q$ , $R$ to $S$ , and $T$ to $U$ , to form triangles. The value of $\angle P + \angle Q + \angle R + \angle S + \angle T + \angle U$ is

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$\text{ (A) }\ 450^{\circ} \qquad\text{ (B) }\ 270^{\circ} \qquad\text{ (C) }\ 360^{\circ} \qquad\text{ (D) }\ 540^{\circ} \qquad\text{ (E) }\ 720^{\circ}$

Solution 1

Using the diagram and the fact that the sum of the interior angles of any triangle is $180^{\circ}$ , We see:

$m \angle POQ = 180^{\circ} - (m \angle P + m \angle Q)$

$m \angle ROS = 180^{\circ} - (m \angle R + m \angle S)$

$m \angle UOT = 180^{\circ} - (m \angle T + m \angle U)$

We also see that angles $ROS$ and $POU$ are vertically opposite, making them equal to each other.

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We see that angles $POQ$ , $POU$ , and $UOT$ form a line. We then have:

$m\angle POQ + m \angle POU + m\angle UOT = 180^{\circ}$

$180^{\circ} - (m \angle P + m\angle Q) + 180^{\circ} - (m \angle R + m \angle S) + 180^{\circ} - (m \angle T + m\angle U) = 180^{\circ}$

Re-arranging terms and adding the $180^{\circ}$ s up, we get:

$540^{\circ} - (m \angle P + m\angle Q + m\angle R + m\angle S + m\angle T + m\angle U) = 180^{\circ}$

We can let $x$ be equal to the value in the parenthesis, which is the value we are trying to find.

$540^{\circ} - x = 180^{\circ}$

$x + 180^{\circ} = 540^{\circ}$

$x = \boxed {\textbf {(C) } 360^{\circ}}$

~anabel.disher

2001 CEMC Gauss (Grade 8) (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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2001 CEMC Gauss (Grade 8) Problems/Problem 21

Problem

Solution 1