Triangle ABC is a right triangle with \angle BAC = 90^\circ and \angle B > \angle C. Suppose \overline{AP} is an altitude of the triangle, \overline{AQ} is an angle bisector of the triangle, and \overline{AR} is a median of the triangle, and \angle PAQ = 13^\circ. If P is on \overline{BQ}, then what is the measure of \angle RAC
2
Let M, N, and P be the midpoints of sides \overline{TU}, \overline{US}, and \overline{ST} of triangle STU, respectively. Let \overline{UZ} be an altitude of the triangle. If \angle TSU = 71^\circ, \angle STU = 36^\circ, and \angle TUS = 73^\circ, then what is \angle NZM + \angle NPM in degrees?
3
Points M, N, and O are the midpoints of sides \overline{KL}, \overline{LJ}, and \overline{JK}, respectively, of triangle JKL. Points P, Q, and R are the midpoints of \overline{NO}, \overline{OM}, and \overline{MN}, respectively. If the area of triangle PQR is 27, then what is the area of triangle LPQ?
4
Point D is the midpoint of median \overline{AM} of triangle ABC. Point E is the midpoint of \overline{AB}, and point T is the intersection of \overline{BD} and \overline{ME}. Find the area of triangle DMT if [ABC] =180.
1) Let's call the measures of angles B and C, B and C, respectively. Since \angle BAC = 90^\circ, we know that \angle B + \angle C = 90^\circ. We also know that \angle PAQ = 13^\circ. Since AP is an altitude and AQ is an angle bisector, this means that \angle BAP = \angle CAQ = 90^\circ - \angle PAQ = 90^\circ - 13^\circ = 77^\circ. Since \angle BAP = \angle CAQ, this means that triangles BAP and CAQ are similar by the Angle-Angle Similarity Postulate (AA~). From this similarity, we can conclude that \angle C = \angle BAC = \angle PAB. Since \angle PAB is an angle in triangle PAB, and \angle CAB is an angle in triangle CAB, we know that \angle CAB + \angle PAB = 180^\circ. Thus, \angle CAB + \angle BAC = 180^\circ, so the three angles in triangle ABC add up to 180^\circ. This means that \angle C = 180^\circ - \angle BAC - \angle B = 180^\circ - 90^\circ - \angle B = 90^\circ - \angle B. Since \angle C = \angle BAC = 90^\circ - \angle B, this means that \angle C + \angle B = 90^\circ - \angle B + \angle B = 90^\circ. We already know that \angle B + \angle C = 90^\circ, so this means that \angle B = \angle C. However, the problem states that \angle B > \angle C, which is a contradiction. Therefore, there is no solution to the problem as stated.
2) Let's call the measure of \angle NZM x degrees and the measure of \angle NPM y degrees. Since M, N, and P are midpoints of sides TU, US, and ST, respectively, this means that triangle SUT is similar to triangle NPQ by the Midline Theorem. From this similarity, we know that \angle NPQ = \angle SUT = 71^\circ, and \angle NQP = \angle TSU = 36^\circ. Since the measures of angles in triangle NPQ add up to 180^\circ, this means that \angle N = 180^\circ - \angle NPQ - \angle NQP = 180^\circ - 71^\circ - 36^\circ = 73^\circ. Similarly, in triangle SMP, we have \angle SMP = \angle SUT = 71^\circ and \angle MSP = \angle STU = 73^\circ. Again, using the fact that the measures of angles in triangle SMP add up to 180^\circ, we have \angle M = 180^\circ - \angle SMP - \angle MSP = 180^\circ - 71^\circ - 73^\circ = 36^\circ. Since \angle M = \angle N, this means that \angle NZM = \angle NPM = x degrees. The problem asks for the sum of the measures of \angle NZM and \angle NPM, which is 2x degrees. Therefore, the answer is \boxed{2x} degrees.
3) Since points P, Q, and R are midpoints of the segments NO, OM, and MN, respectively, this means that triangle PQR is similar to triangle LNK by the Midline Theorem. Since the Midline Theorem only states that the midline of a triangle is parallel to the third side and half the length of the third side, we have no information about the orientation of triangle PQR with respect to triangle LNK. Therefore, without more information, we cannot determine the area of triangle LPQ.
4) Since E is the midpoint of AB, this means that \frac{AE}{EB} = 1. Therefore, if D is the midpoint of AM, this means that \frac{AD}{DM} = 1. Since T is the intersection of BD and ME, this means that \frac{TE}{EB} = \frac{TM}{MD}. Thus, \frac{TM}{MD} = \frac{TE}{EB} = \frac{1}{1} = 1. Therefore, TM = MD. Since T is the midpoint of segment MD, this means that triangle DMT is an isosceles right triangle. Since the median of a right triangle is half the length of the hypotenuse, this means that DT = TM = \frac{1}{2}DM. Since the area of triangle ABC is 180, this means that the area of triangle DMT is \frac{1}{4} the area of triangle ABC, or \frac{1}{4}(180) = \boxed{45}.