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Difference between revisions of "2001 CEMC Gauss (Grade 8) Problems/Problem 21"

(Created page with "==Problem== What number should be placed in the box to make <math>10^{4} \times 100^{\Box} = 1000^{6}</math>? <math> \text{ (A) }\ 450^{\circ} \qquad\text{ (B) }\ 270^{\cir...")
 
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==Problem==   
 
==Problem==   
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What number should be placed in the box to make <math>10^{4} \times 100^{\Box} = 1000^{6}</math>?
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Lines <math>PS</math>, <math>QT</math> and <math>RU</math> intersect at a common point <math>O</math>, as shown. <math>P</math> is joined to <math>Q</math>, <math>R</math> to <math>S</math>, and <math>T</math> to <math>U</math>, to form triangles. The value of <math>\angle P + \angle Q + \angle R + \angle S + \angle T + \angle U</math> is
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{{Image needed}}
 
<math> \text{ (A) }\ 450^{\circ} \qquad\text{ (B) }\ 270^{\circ} \qquad\text{ (C) }\ 360^{\circ} \qquad\text{ (D) }\ 540^{\circ} \qquad\text{ (E) }\ 720^{\circ}</math>   
 
<math> \text{ (A) }\ 450^{\circ} \qquad\text{ (B) }\ 270^{\circ} \qquad\text{ (C) }\ 360^{\circ} \qquad\text{ (D) }\ 540^{\circ} \qquad\text{ (E) }\ 720^{\circ}</math>   
 
==Solution 1==
 
==Solution 1==

Latest revision as of 20:29, 22 October 2025

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

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

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

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.

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

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)
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Problem 22
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CEMC Gauss (Grade 8)
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