In the diagram, $P$ is on $RS$ so that $QP$ bisects $\angle SQR$.  Also, $PQ=PR$, $\angle RSQ=2y^\circ$, and $\angle RPQ=3y^\circ$.  What is the measure, in degrees, of $\angle RPQ$? [asy]
// C14
import olympiad;
size(7cm);

real x = 50; real y = 20;

pair q = (1, 0);
pair r = (0, 0);
pair p = intersectionpoints((10 * dir(x))--r, q--(shift(q) * 10 * dir(180 - x)))[0];
pair s = intersectionpoints(r--(r + 10 * (p - r)), 10 * dir(180 - 2 * x)--q)[0];

// Draw lines
draw(p--s--q--p--r--q);

// Label points
label("$R$", r, SW);
label("$Q$", q, SE);
label("$S$", s, N);
label("$P$", p, NW);

// Label angles
label("$x^\circ$", q, 2 * W + 2 * NW);
label("$x^\circ$", q, 4 * N + 2 * NW);
label("$2y^\circ$", s, 5 * S + 4 * SW);
label("$3y^\circ$", p, 4 * S);

// Tick marks
add(pathticks(r--p, 2, spacing=0.6, s=2));
add(pathticks(p--q, 2, spacing=0.6, s=2));
[/asy]
Since $RPS$ is a straight line, then $\angle SPQ = 180^\circ - \angle RPQ = 180^\circ - 3y^\circ$.

Using the angles in $\triangle PQS$, we have $\angle PQS + \angle QSP + \angle SPQ = 180^\circ$. Thus, $x^\circ+2y^\circ + (180^\circ - 3y^\circ) = 180^\circ$ or $x-y+180 = 180$ or $x=y$.

(We could have instead looked at $\angle RPQ$ as being an external angle to $\triangle SPQ$.)

Since $x=y$, then $\angle RQS=2y^\circ$.

Since $RP=PQ$, then $\angle PRQ=\angle PQR=x^\circ = y^\circ$. [asy]
// C16S
import olympiad;
size(7cm);

real x = 36; real y = 36;

pair q = (1, 0);
pair r = (0, 0);
pair p = intersectionpoints((10 * dir(x))--r, q--(shift(q) * 10 * dir(180 - x)))[0];
pair s = intersectionpoints(r--(r + 10 * (p - r)), 10 * dir(180 - 2 * x)--q)[0];

// Draw lines
draw(p--s--q--p--r--q);

// Label points
label("$R$", r, SW);
label("$Q$", q, SE);
label("$S$", s, N);
label("$P$", p, NW);

// Label angles
label("$y^\circ$", q, 4 * W + 2 * NW);
label("$y^\circ$", q, N + 5 * NW);
label("$y^\circ$", r, 2 * NE + 3 * E);
label("$2y^\circ$", s, 3 * S + SW);
label("$3y^\circ$", p, 3 * S);

// Tick marks
add(pathticks(r--p, 2, spacing=0.6, s=2));
add(pathticks(p--q, 2, spacing=0.6, s=2));

[/asy] Therefore, the angles of $\triangle RQS$ are $y^\circ$, $2y^\circ$ and $2y^\circ$.

Thus, $y^\circ+2y^\circ+2y^\circ=180^\circ$ or $5y=180$ or $y=36$.

Therefore, $\angle RPQ=3y^\circ = 3(36)^\circ=108^\circ$, so our final answer is $\boxed{108}$ degrees.
108
Level 3
Geometry
Problem: In the diagram, $P$ is on $RS$ so that $QP$ bisects $\angle SQR$.  Also, $PQ=PR$, $\angle RSQ=2y^\circ$, and $\angle RPQ=3y^\circ$.  What is the measure, in degrees, of $\angle RPQ$? [asy]
// C14
import olympiad;
size(7cm);

real x = 50; real y = 20;

pair q = (1, 0);
pair r = (0, 0);
pair p = intersectionpoints((10 * dir(x))--r, q--(shift(q) * 10 * dir(180 - x)))[0];
pair s = intersectionpoints(r--(r + 10 * (p - r)), 10 * dir(180 - 2 * x)--q)[0];

// Draw lines
draw(p--s--q--p--r--q);

// Label points
label("$R$", r, SW);
label("$Q$", q, SE);
label("$S$", s, N);
label("$P$", p, NW);

// Label angles
label("$x^\circ$", q, 2 * W + 2 * NW);
label("$x^\circ$", q, 4 * N + 2 * NW);
label("$2y^\circ$", s, 5 * S + 4 * SW);
label("$3y^\circ$", p, 4 * S);

// Tick marks
add(pathticks(r--p, 2, spacing=0.6, s=2));
add(pathticks(p--q, 2, spacing=0.6, s=2));
[/asy]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the measure of angle RPQ, we start by noting that angle QP bisects angle SQR. This means that angle SQR is divided into two equal parts, each measuring x degrees. Therefore, we can express angle SQR as 2x degrees.

Next, we know that the sum of angles in triangle SQR must equal 180 degrees. Thus, we can set up the equation:

angle SQR + angle RSQ + angle QSR = 180 degrees.

Substituting the known values, we have:

2x + 2y + x = 180 degrees.

This simplifies to:

3x + 2y = 180 degrees. (1)

We also know that PQ = PR, which indicates that triangle PQR is isosceles. In an isosceles triangle, the angles opposite the equal sides are equal. Therefore, angle PQR is equal to angle PRQ, both measuring x degrees.

Now, we can express the sum of angles in triangle PQR:

angle RPQ + angle PQR + angle PRQ = 180 degrees.

Substituting the known values, we have:

angle RPQ + x + x = 180 degrees.

This simplifies to:

angle RPQ + 2x = 180 degrees. (2)

From equation (2), we can express angle RPQ as:

angle RPQ = 180 degrees - 2x. (3)

Next, we need to relate y and x. We know from the problem that angle RSQ = 2y degrees. Since angle SQR is 2x degrees, we can express the relationship between y and x using equation (1):

3x + 2y = 180 degrees.

To isolate y, we can rearrange this equation:

2y = 180 degrees - 3x,
y = (180 degrees - 3x) / 2. (4)

Now, we substitute equation (4) into the expression for angle RPQ. We need to find a value for y in terms of x to solve for angle RPQ. 

We also know that angle RPQ is given as 3y degrees. Substituting equation (4) into this gives:

angle RPQ = 3 * (180 degrees - 3x) / 2,
angle RPQ = (540 degrees - 9x) / 2.

Now we have two expressions for angle RPQ: one from equation (3) and one from the substitution. Setting them equal gives:

180 degrees - 2x = (540 degrees - 9x) / 2.

To eliminate the fraction, we multiply through by 2:

2(180 degrees - 2x) = 540 degrees - 9x,
360 degrees - 4x = 540 degrees - 9x.

Rearranging this equation leads to:

5x = 180 degrees,
x = 36 degrees.

Now substituting x back into equation (3) to find angle RPQ:

angle RPQ = 180 degrees - 2(36 degrees),
angle RPQ = 180 degrees - 72 degrees,
angle RPQ = 108 degrees.

Thus, the measure of angle RPQ is 108 degrees.<|END_OF_LATENT|> The final answer is $108$.