Question 19·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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Two lines intersect at exactly one point, forming two acute angles and two obtuse angles. The measure of one of the acute angles is $\left(\frac{5}{3}x - 40\right)^\circ$ .

A third line passes through the intersection point and bisects each of the obtuse angles formed by the two original lines. A fourth line passes through the intersection point and bisects each of the acute angles formed by the two original lines.

Which choice is NOT equivalent (as an expression in $x$ ) to the sum of the measures of any three of the eight angles formed by the four lines?

Intersecting lines make a linear pair: a neighboring acute and obtuse angle total $180^\circ$ . After both types are bisected, count how many of the three chosen pieces come from each type. Write the possible sums in $x$ before comparing choices. The trap is treating a plus sign on an $x$ term as interchangeable with a minus sign.

Hints

Hint 1
An angle bisector cuts an angle into two equal parts. Both original acute angles are bisected, so four pieces each have half the measure of the given acute angle.
Hint 2
A neighboring acute and obtuse angle form a linear pair, totaling $180^\circ$ . If you halve both angles, what must one acute piece plus one obtuse piece total?
Hint 3
Two equivalent expressions agree at every allowed $x$ . List three-piece sums by using zero, one, two, or three acute pieces. One valid input can disprove a match, but one match cannot prove equivalence.

Step-by-step

Count the angle pieces
Step 1
Name the given acute angle
Call its measure $a(x)$ . Type $a(x)=\frac{5}{3}x-40$ , then $a(60)$ . Desmos prints $60$ , so $x=60$ gives a valid acute angle. We'll use that input to check a choice later.
Step 2
Find an acute piece
An angle bisector divides an angle into equal halves. Both acute angles are bisected, making four equal acute pieces. Substitute the given measure to call each piece $s(x)$ :
$s(x)=\frac{\frac{5}{3}x-40}{2}.$
Divide each term by $2$ :
$s(x)=\frac{5}{6}x-20.$
Type $s(x)=a(x)/2$ and $s(60)$ ; Desmos prints $30$ .
Step 3
Relate an obtuse piece to an acute piece
A neighboring acute and obtuse angle form a linear pair, totaling $180^\circ$ . Halving both leaves pieces totaling $90^\circ$ . So each of the four obtuse pieces measures:
$t(x)=90-s(x).$
Type $t(x)=90-s(x)$ and $t(60)$ ; Desmos prints $60$ .
Step 4
List every type of three-piece sum
Three chosen pieces can include zero, one, two, or three acute pieces. Their sums are $3t(x)$ , $s(x)+2t(x)$ , $2s(x)+t(x)$ , and $3s(x)$ . Their positions around the intersection don't change their measures, so these cover any three of the eight angles. Count pieces by type before comparing expressions.
Step 5
Find the sum of three acute pieces
Substitute the measure of an acute piece:
$3s(x)=3\left(\frac{5}{6}x-20\right).$
Distribute $3$ :
$3s(x)=\frac{5}{2}x-60.$
So the expression with $-60$ is a possible sum.
Step 6
Find the sum with two acute pieces
Replace the one obtuse piece with $90-s(x)$ :
$2s(x)+t(x)=2s(x)+90-s(x).$
Combine the $s(x)$ terms:
$2s(x)+t(x)=90+s(x).$
Insert the measure of $s(x)$ :
$2s(x)+t(x)=90+\frac{5}{6}x-20.$
Combine constants:
$2s(x)+t(x)=70+\frac{5}{6}x.$
This is another possible sum.
Step 7
Find the sum with one acute piece
Replace both obtuse pieces:
$s(x)+2t(x)=s(x)+2(90-s(x)).$
Distribute $2$ :
$s(x)+2t(x)=s(x)+180-2s(x).$
Combine the $s(x)$ terms:
$s(x)+2t(x)=180-s(x).$
Insert the measure of $s(x)$ :
$s(x)+2t(x)=180-\left(\frac{5}{6}x-20\right).$
Distribute the minus sign:
$s(x)+2t(x)=180-\frac{5}{6}x+20.$
Add constants:
$s(x)+2t(x)=200-\frac{5}{6}x.$
The minus sign on the $x$ term matters.
Step 8
Find the remaining three-piece sum
Replace each obtuse piece with $90-s(x)$ :
$3t(x)=3(90-s(x)).$
Distribute $3$ :
$3t(x)=270-3s(x).$
Insert the three-acute-piece sum found above:
$3t(x)=270-\left(\frac{5}{2}x-60\right).$
Distribute the minus sign:
$3t(x)=270-\frac{5}{2}x+60.$
Add constants:
$3t(x)=330-\frac{5}{2}x.$
This last possible sum has a negative, not positive, $x$ term.
Step 9
Rule out the expression that does not match
At the valid input $x=60$ , each acute piece is $30^\circ$ and each obtuse piece is $60^\circ$ . So the largest sum of three pieces is $3t(60)$ . Type $3t(60)$ and $330+\frac{5}{2}(60)$ ; Desmos prints $180$ and $480$ . Expressions that differ at even one allowed input aren't equivalent. So $\left(330+\frac{5}{2}x\right)^\circ$ cannot be the sum of any three angles. Choice D.

Question 19·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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