AMC 10 · 2025 · #13
Grade 8 geometry-2dThe altitude to the hypotenuse of a 30∘−60∘−90∘ is divided into two segments of lengths x<y by the median to the shortest side of the triangle. What is the ratio x+yx?
Pick an answer.
AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: In a 30-60-90 right triangle, drop the altitude from the right angle onto the hypotenuse. Also draw the median from the far vertex to the midpoint of the shortest side. This median crosses the altitude and cuts it into two pieces, a shorter one x and a longer one y. Find x/(x+y).
Givens: The triangle has angles 30, 60, and 90 degrees.; One segment is the altitude drawn from the right angle to the hypotenuse.; The other segment is the median drawn to the shortest side (from the opposite vertex to that side's midpoint).; The median splits the altitude into two lengths x and y with x < y.
Unknowns: The ratio x/(x+y), where x and y are the two pieces of the altitude.
Understand
Restated: In a 30-60-90 right triangle, drop the altitude from the right angle onto the hypotenuse. Also draw the median from the far vertex to the midpoint of the shortest side. This median crosses the altitude and cuts it into two pieces, a shorter one x and a longer one y. Find x/(x+y).
Givens: The triangle has angles 30, 60, and 90 degrees.; One segment is the altitude drawn from the right angle to the hypotenuse.; The other segment is the median drawn to the shortest side (from the opposite vertex to that side's midpoint).; The median splits the altitude into two lengths x and y with x < y.
Plan
Primary tool: #1 Draw a Diagram
Secondary: #13 Convert to Algebra, #7 Identify Subproblems, #4 Introduce a Variable
The problem is pure position and shape, so the safest move is to draw it and pin it to a grid. Put the right angle at the origin so the two legs lie on the axes, and the altitude, the median, and their crossing point all become lines you can write with equations. Then the whole question turns into finding where two lines meet and reading off how that point splits the altitude. No clever trick is needed once the picture is on coordinates.
Execute — Answer: A
7.G.A.2 Step 1 Name the parts of the triangle
- Call the right-angle vertex B, the 60 degree vertex A, and the 30 degree vertex C.
- The shortest side is the leg opposite the 30 degree angle, which is AB.
- So the median goes from C to the midpoint of AB, and the altitude drops from B straight down onto the hypotenuse AC.
💡 Labeling each vertex by its angle keeps the shortest side, the altitude, and the median from getting mixed up.
8.G.B.7 Step 2 Find the 30-60-90 side lengths
- Cut an equilateral triangle of side 2 in half.
- The base splits into 1, and the Pythagorean theorem gives the height as the square root of (2 squared minus 1 squared), which is the square root of 3.
- So the sides opposite 30, 60, 90 degrees are in the ratio 1 to root 3 to 2.
- Doubling for whole numbers, take AB = 2, BC = 2 root 3, and hypotenuse AC = 4.
💡 Half of an equilateral triangle is exactly a 30-60-90, so the Pythagorean theorem hands you the root 3.
6.NS.C.6 Step 3 Place the triangle on a grid
- Put the right angle B at the origin and lay the two legs along the axes.
- Then B = (0, 0), A = (2, 0) along the x-axis, and C = (0, 2 root 3) up the y-axis.
- Now every point in the problem has coordinates, so lines can be written as equations.
💡 Standing the right angle on the origin lets the two legs ride the axes, so coordinates fall out for free.
8.EE.B.6 Step 4 Locate the foot of the altitude
- The hypotenuse AC runs from (2, 0) to (0, 2 root 3), so its slope is negative root 3 and its line is y = -root 3 times x + 2 root 3.
- The altitude from B is perpendicular to it, so its slope is 1 over root 3, giving the line y = x over root 3.
- Setting these equal solves for the foot D: x = 3/2 and y = root 3 over 2, so D = (3/2, root 3 over 2).
💡 Perpendicular slopes are negative reciprocals, so the altitude's equation writes itself once you have the hypotenuse's.
6.G.A.3 Step 5 Write the median as a line
- The midpoint of the shortest side AB is the average of A = (2, 0) and B = (0, 0), which is E = (1, 0).
- The median joins C = (0, 2 root 3) to E = (1, 0).
- Its slope is negative 2 root 3, so its line is y = -2 root 3 times x + 2 root 3.
💡 A midpoint is just the average of the two endpoints, so the median's endpoints are easy to read off.
8.EE.C.8 Step 6 Cross the two lines and split the altitude
- Find where the median CE meets the altitude BD by solving y = x over root 3 together with y = -2 root 3 times x + 2 root 3.
- This gives x = 6/7, so the crossing point F sits 6/7 of the way in the x-direction while D sits at 3/2.
- That makes F land 4/7 of the distance from B to D, leaving 3/7 from F to D.
- So the piece by the hypotenuse is DF = 3/7 of BD and the other piece is BF = 4/7 of BD.
- The shorter piece is x = DF, so x over (x + y) = (3/7) over 1 = 3/7.
- The answer is (A).
💡 Where two lines meet is one point that satisfies both equations, and that point's place along the altitude is the whole answer.
7.G.A.2 Call the right-angle vertex B, the 60 degree vertex A, and the 30 degree vertex 8.G.B.7 Cut an equilateral triangle of side 2 in half. The base splits into 1, and the P 6.NS.C.6 Put the right angle B at the origin and lay the two legs along the axes. Then B 8.EE.B.6 The hypotenuse AC runs from (2, 0) to (0, 2 root 3), so its slope is negative ro 6.G.A.3 The midpoint of the shortest side AB is the average of A = (2, 0) and B = (0, 0) 8.EE.C.8 Find where the median CE meets the altitude BD by solving y = x over root 3 toge Review
Reasonableness: The crossing point F = (6/7, 2 root 3 over 7) lies on both lines: 6/7 over root 3 equals 2 root 3 over 7, and -2 root 3 times 6/7 plus 2 root 3 also equals 2 root 3 over 7. Since F is closer to the hypotenuse (D) than to the right angle (B), the piece DF really is the shorter one, matching x < y. The ratio 3/7 is a little under one half, which fits a picture where the shorter piece is only slightly smaller than the longer piece. Answer (A) holds.
Alternative: Use mass points. Because the altitude foot divides the hypotenuse so that AD = 1 and DC = 3, hang weight 3 at A and weight 1 at C, making the foot weigh 4. The median goes to the midpoint of AB, which forces weight 3 at B as well. On the altitude BD, the crossing point balances weight 3 at B against weight 4 at D, so it splits BD in the ratio DF to FB = 3 to 4. Then x/(x+y) = 3/(3+4) = 3/7, the same answer with no coordinates.
CCSS standards used (min grade 8)
7.G.A.2Draw geometric shapes with given conditions (Drawing and labeling the 30-60-90 triangle and marking its altitude, median, and shortest side.)8.G.B.7Apply the Pythagorean Theorem to find unknown side lengths in right triangles (Deriving the 1 : root 3 : 2 side ratio by splitting an equilateral triangle and computing the height.)6.NS.C.6Understand and plot points in the coordinate plane (Placing the right angle at the origin and assigning coordinates to the three vertices.)8.EE.B.6Derive and use the equation y = mx + b for a line, including slope (Writing the hypotenuse and the perpendicular altitude as line equations to find the foot D.)6.G.A.3Draw polygons in the coordinate plane and use coordinates to find lengths (Finding the midpoint of the shortest side and setting up the median as a segment between two points.)8.EE.C.8Solve systems of two linear equations by finding their intersection (Solving the altitude and median equations together to locate F and read off how it splits the altitude.)
⭐ When a shape question is all about where lines cross, drop it onto a grid and let the equations find the meeting point for you.
⭐ When a shape question is all about where lines cross, drop it onto a grid and let the equations find the meeting point for you.
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