Question 121·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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In the figure, $PQ$ is the diameter of a circle. Point $R$ lies on the circle, and segment $RS$ is perpendicular to $PQ$ at point $S$ .

If $PQ=130$ and $PR=13\sqrt{5}$ , which choice is the value of $\frac{PQ}{PS}$ ?

Note: Figure not drawn to scale.

Joining a point on a circle to both ends of a diameter makes a right triangle. A perpendicular from that point to the diameter creates a smaller triangle with the same angles, so corresponding sides give a proportion. Use it to find the needed piece of the diameter, then take the ratio the question asks for. The piece itself isn't the answer.

Hints

Hint 1
An inscribed angle has its vertex on a circle. If its sides reach the two ends of a diameter, it measures $90^\circ$ . Where is that angle in the large triangle?
Hint 2
Two triangles are similar when their matching angles are equal. The large triangle and the smaller triangle near $P$ share the angle at $P$ and each has a right angle. Which sides match?
Hint 3
The proportion can give you $PS$ , but the target is $PQ$ divided by $PS$ . Keep that last division in view.

Step-by-step

Use the two similar triangles
Step 1
Find the right angle in the large triangle
No Desmos needed yet. First read the circle geometry. An inscribed angle has its vertex on the circle. Because $R$ lies on the circle and $PQ$ is a diameter, angle $PRQ$ spans a diameter, so it measures $90^\circ$ . So $PQ$ is the hypotenuse, the side opposite the right angle, of triangle $PQR$ .
Step 2
Match sides of the similar triangles
Since $RS$ is perpendicular to $PQ$ at $S$ , triangle $PSR$ is right at $S$ . It shares the angle at $P$ with triangle $PQR$ , so the triangles are similar: their matching angles are equal. Their hypotenuses match, $PR$ with $PQ$ ; the sides beside angle $P$ match, $PS$ with $PR$ . So their ratios are equal:
$\frac{PS}{PR}=\frac{PR}{PQ}.$
Step 3
Relate the given leg to the needed piece
Multiply both sides of the proportion by $PR\cdot PQ$ : $PR^2=PQ\cdot PS$ .A leg squared equals the whole hypotenuse times the piece beside that leg. Here $PR$ touches the piece $PS$ , not the piece near $Q$ .
Step 4
Find the length of $PS$
Use the given $PR=13\sqrt{5}$ and $PQ=130$ . Type $(13\sqrt{5})^2\sim130x_1$ in Desmos, with $x_1$ standing for $PS$ . The $\sim$ tells Desmos to find the value that fits the equation. Under PARAMETERS, it shows $x_1=6.5$ , so $PS=6.5$ .
Step 5
Take the requested ratio
The $6.5$ is $PS$ , not the answer. Keep the first Desmos line and type $130/x_1$ ; Desmos displays $20$ . So $\frac{PQ}{PS}=20$ . Choice C.

Lessons that teach this

SAT Geometry and TrigonometryAdvancedCoreUse similarity inside right triangles

Question 121·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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Strategy

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Step-by-step

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