AMC 10 · 2025 · #13

Grade 8 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Line Intersection Vertices then Area

thirty-sixty-ninety-trianglesimilar-trianglescentroid-2-to-1 identify-subproblems ↑ Prerequisites: similar-triangles

📏 Long solution 💡 4 insights

Problem

In a 30-60-90 right triangle, drop the altitude from the right angle onto the hypotenuse. Also draw the median to the shortest side, the segment joining the opposite vertex to the midpoint of that side. This median crosses the altitude and splits it into a shorter piece

x

and a longer piece

y

. Find the ratio

\frac{x}{x+y}

Pick an answer.

(A)

$\dfrac{3}{7}$

(B)

$\dfrac{\sqrt3}{4}$

(C)

$\dfrac{4}{9}$

(D)

$\dfrac{5}{11}$

(E)

$\dfrac{4\sqrt3}{15}$

AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

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.

1STEP 1

Name the parts of the triangle

Call the right angle $B$ , the 60 degree vertex $A$ , the 30 degree vertex $C$ . The short side is $AB$ , so the median runs from $C$ to its midpoint.

∠ B = 90°, ∠ A = 60°, ∠ C = 30°; shortest side = AB

2STEP 2

Find the 30-60-90 side lengths

Half an equilateral triangle is a 30-60-90, giving sides in ratio $1:\sqrt{3}:2$ , so take $AB=2$ , $BC=2\sqrt{3}$ , $AC=4$ .

1:√3:2 → AB=2, BC=2√3, AC=4 (2²+(2√3)²=4+12=16=4²)

3STEP 3

Place the triangle on a grid

Put $B$ at the origin with both legs on the axes: $B=(0,0)$ , $A=(2,0)$ , $C=(0,2\sqrt{3})$ . Now every line can be written as an equation.

B=(0,0), A=(2,0), C=(0,2√3)

4STEP 4

Locate the foot of the altitude

Cross $AC: y=-\sqrt{3}x+2\sqrt{3}$ with the perpendicular $BD: y=\frac{x}{\sqrt{3}}$ to get the foot $D=(\frac{3}{2},\frac{\sqrt{3}}{2})$ .

AC: y=-√3 x+2√3, BD: y=x/√3 → x/√3=-√3 x+2√3 → x=3/2, D=(3/2,√3/2)

5STEP 5

Write the median as a line

The midpoint of $AB$ averages to $E=(1,0)$ , so the median from $C$ is the line $CE: y=-2\sqrt{3}x+2\sqrt{3}$ .

E=((2+0)/2,(0+0)/2)=(1,0), CE: y=-2√3 x+2√3

6STEP 6

Cross the two lines and split the altitude

The lines meet at horizontal position $\frac{6}{7}$ while $D$ sits at $\frac{3}{2}$ , so the piece by the hypotenuse is $\frac{3}{7}$ of the altitude.

x/√3=-2√3 x+2√3→ x=6/7; BF/BD=6/7/3/2=4/7, DF/BD=3/7 → x/(x+y)=3/7

Answer

3/7

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.

💡Key takeaway

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.

Name the parts of the triangle
Find the 30-60-90 side lengths
Place the triangle on a grid
Locate the foot of the altitude
Write the median as a line
Cross the two lines and split the altitude