AMC 10 · 2011 · #18
Grade 8 geometry-2dRectangle ABCD has AB=6 and BC=3. Point M is chosen on side AB so that ∠AMD=∠CMD. What is the degree measure of ∠AMD?
Pick an answer.
AMC 10 2011 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 rectangle $ABCD$ the long side $AB = 6$ and the short side $BC = 3$. A point $M$ is placed on side $AB$ so that segment $MD$ splits the corner it makes with the sides equally: $\angle AMD = \angle CMD$. Find the degree measure of $\angle AMD$.
Givens: $ABCD$ is a rectangle, so all four corners are right angles and opposite sides are equal; $AB = CD = 6$ and $BC = AD = 3$; $M$ lies on side $AB$; $\angle AMD = \angle CMD$ (the segment $MD$ makes equal angles on its two sides); Answer choices: (A) $15$, (B) $30$, (C) $45$, (D) $60$, (E) $75$
Unknowns: The degree measure of $\angle AMD$
Understand
Restated: In rectangle $ABCD$ the long side $AB = 6$ and the short side $BC = 3$. A point $M$ is placed on side $AB$ so that segment $MD$ splits the corner it makes with the sides equally: $\angle AMD = \angle CMD$. Find the degree measure of $\angle AMD$.
Givens: $ABCD$ is a rectangle, so all four corners are right angles and opposite sides are equal; $AB = CD = 6$ and $BC = AD = 3$; $M$ lies on side $AB$; $\angle AMD = \angle CMD$ (the segment $MD$ makes equal angles on its two sides); Answer choices: (A) $15$, (B) $30$, (C) $45$, (D) $60$, (E) $75$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #4 Introduce a Variable
No picture is given, so the first move is to draw the rectangle and mark the two equal angles at $M$ (Tool #1). The drawing exposes two facts you would otherwise miss: $AB \parallel DC$ turns $\angle AMD$ into an alternate interior angle equal to $\angle MDC$, and that makes triangle $CMD$ isosceles. From there Tool #7 (Identify Subproblems) splits the work into a short isosceles-triangle step and a right-triangle step, and Tool #4 (Introduce a Variable) lets the straight-line angle sum finish it in one equation.
Execute — Answer: E
4.G.A.1 Step 1 Draw and label the rectangle
- Put the rectangle with $A$ bottom-left, $B$ bottom-right, $C$ top-right, $D$ top-left, so $AB = 6$ is the bottom and $BC = 3$ is the right side.
- Place $M$ somewhere on the bottom edge $AB$ and draw $MD$ and $MC$.
- Mark the two equal angles $\angle AMD$ and $\angle CMD$.
- The point of a diagram is to make the parallel sides and the shared segment $MD$ visible.
💡 A labeled picture turns hidden geometry conditions into lines and angles you can actually chase.
8.G.A.5 Step 2 Spot the alternate interior angles
- The top side $DC$ is parallel to the bottom side $AB$, and $MD$ is a transversal crossing both.
- So $\angle AMD$ (at $M$, below) and $\angle MDC$ (at $D$, above) are alternate interior angles, which are equal.
💡 When two parallel lines are cut by a slanted line, the two inside angles on opposite sides are twins.
8.G.A.5 Step 3 Find the isosceles triangle
- We are given $\angle AMD = \angle CMD$, and we just found $\angle AMD = \angle MDC$.
- Chaining these, $\angle CMD = \angle MDC$.
- In triangle $CMD$ the angle at $M$ equals the angle at $D$, so the sides opposite them are equal: $CM = CD$.
- Since $CD = AB = 6$, we get $CM = 6$.
💡 Equal base angles force equal sides, so a triangle with two matching angles is isosceles.
8.G.B.7 Step 4 Read the right triangle at B
- Look at triangle $CMB$.
- The corner at $B$ is a right angle, the leg $CB = 3$, and the hypotenuse is $CM = 6$.
- By the Pythagorean Theorem the other leg is $MB = \sqrt{6^2 - 3^2} = \sqrt{27} = 3\sqrt{3}$, so the sides are $3 : 3\sqrt{3} : 6$, the classic $1:\sqrt{3}:2$ shape.
- A right triangle whose short leg is exactly half the hypotenuse is a $30$-$60$-$90$ triangle, and the angle facing the short leg $CB$ is $\angle CMB = 30^\circ$.
💡 A right triangle with a leg half the hypotenuse is always the 30-60-90 triangle.
4.MD.C.7 Step 5 Add the angles along the line
- The three angles at $M$ sit along the straight edge $AB$, so they add to $180^\circ$: $\angle AMD + \angle DMC + \angle CMB = 180^\circ$.
- Let $\theta = \angle AMD$; since $\angle DMC = \angle AMD = \theta$ and $\angle CMB = 30^\circ$, this becomes $2\theta + 30^\circ = 180^\circ$.
- Solving gives $\theta = 75^\circ$.
- So $\angle AMD = 75^\circ$, the answer is (E).
💡 Angles resting on one straight line always total a straight angle, $180^\circ$.
4.G.A.1 Put the rectangle with $A$ bottom-left, $B$ bottom-right, $C$ top-right, $D$ top 8.G.A.5 The top side $DC$ is parallel to the bottom side $AB$, and $MD$ is a transversal 8.G.A.5 We are given $\angle AMD = \angle CMD$, and we just found $\angle AMD = \angle M 8.G.B.7 Look at triangle $CMB$. The corner at $B$ is a right angle, the leg $CB = 3$, an 4.MD.C.7 The three angles at $M$ sit along the straight edge $AB$, so they add to $180^\c Review
Reasonableness: Check that $\angle AMD = 75^\circ$ holds together. If $\angle CMB = 30^\circ$ then $MB = CB/\tan 30^\circ = 3\sqrt{3} \approx 5.20$, so $M \approx (0.80, 0)$ and $CM = \sqrt{(3\sqrt3)^2 + 3^2} = \sqrt{27+9} = 6 = CD$, confirming the isosceles triangle. The three angles at $M$ are $75^\circ + 75^\circ + 30^\circ = 180^\circ$, exactly a straight line. Everything closes, and $75^\circ$ is choice (E).
Alternative: Use coordinates and trigonometry. With $M=(m,0)$, $\angle AMD$ has $\tan\angle AMD = 3/m$, and requiring $MD$ to bisect $\angle AMC$ forces $CM = 6$, giving $m = 6 - 3\sqrt{3}$. Then $\tan\angle AMD = 3/(6-3\sqrt3) = 2+\sqrt3 = \tan 75^\circ$, so $\angle AMD = 75^\circ$ again. A faster contest shortcut: a quick sketch shows the angle is clearly larger than $60^\circ$, eliminating (A)-(D) and leaving (E).
CCSS standards used (min grade 8)
4.G.A.1Draw points, lines, line segments, rays, angles, and perpendicular and parallel lines (Drawing and labeling the rectangle, placing $M$ on $AB$, and marking the two equal angles and the parallel sides to set up the angle chase.)8.G.A.5Use informal arguments about angle sums and angles formed by parallel lines cut by a transversal (Recognizing $\angle AMD = \angle MDC$ as alternate interior angles across $AB \parallel DC$, then concluding triangle $CMD$ is isosceles with $CM = CD$.)8.G.B.7Apply the Pythagorean Theorem to determine unknown side lengths in right triangles (Finding $MB = 3\sqrt{3}$ in right triangle $CMB$ and recognizing the $1:\sqrt3:2$ ratio, so $\angle CMB = 30^\circ$.)4.MD.C.7Recognize angle measure as additive and find unknown angles by adding and subtracting (Adding the three angles at $M$ along line $AB$ to $180^\circ$ and solving $2\theta + 30^\circ = 180^\circ$ for $\theta = 75^\circ$.)
⭐ Parallel sides made the angle bounce back equal, turning a triangle isosceles; then a 30-60-90 corner and the straight-line total of $180^\circ$ pinned the angle at $75^\circ$.
⭐ Parallel sides made the angle bounce back equal, turning a triangle isosceles; then a 30-60-90 corner and the straight-line total of $180^\circ$ pinned the angle at $75^\circ$.
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