AMC 10 · 2012 · #12
Grade 8 geometry-2dPoint B is due east of point A. Point C is due north of point B. The distance between points A and C is 102, and ∠BAC=45∘. Point D is 20 meters due north of point C. The distance AD is between which two integers?
Pick an answer.
AMC 10 2012 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: Starting from point $A$, go east to $B$, then north to $C$. The straight distance $AC$ is $10\sqrt{2}$ and the angle at $A$ in triangle $ABC$ is $45^\circ$. Point $D$ sits $20$ meters north of $C$ (so $B$, $C$, $D$ lie on one north line). Find which two whole numbers the distance $AD$ falls between.
Givens: $B$ is due east of $A$, and $C$ is due north of $B$; $AC = 10\sqrt{2}$ and $\angle BAC = 45^\circ$; $D$ is $20$ meters due north of $C$; Answer choices name pairs of consecutive integers: (A) $30$ and $31$, (B) $31$ and $32$, (C) $32$ and $33$, (D) $33$ and $34$, (E) $34$ and $35$
Unknowns: The length $AD$, and the two consecutive integers it lies between
Understand
Restated: Starting from point $A$, go east to $B$, then north to $C$. The straight distance $AC$ is $10\sqrt{2}$ and the angle at $A$ in triangle $ABC$ is $45^\circ$. Point $D$ sits $20$ meters north of $C$ (so $B$, $C$, $D$ lie on one north line). Find which two whole numbers the distance $AD$ falls between.
Givens: $B$ is due east of $A$, and $C$ is due north of $B$; $AC = 10\sqrt{2}$ and $\angle BAC = 45^\circ$; $D$ is $20$ meters due north of $C$; Answer choices name pairs of consecutive integers: (A) $30$ and $31$, (B) $31$ and $32$, (C) $32$ and $33$, (D) $33$ and $34$, (E) $34$ and $35$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities
The problem is pure compass directions, so Tool #1 (Draw a Diagram) turns "east" and "north" into a picture where those two directions meet at a right angle at $B$ — that single right angle is the key that unlocks everything. Tool #7 (Identify Subproblems) splits the work into two right triangles that share the leg $AB$: the small triangle $ABC$ pins down the eastward distance $AB$, and the big triangle $ABD$ gives $AD$. Finally Tool #3 (Eliminate Possibilities) squares the boundary integers to trap $AD$ between the correct consecutive pair.
Execute — Answer: B
4.G.A.1 Step 1 Draw the directions as right angles
- Place $A$ at the origin.
- "East" is the horizontal direction and "north" is the vertical direction, and those two directions are perpendicular.
- Since $B$ is east of $A$ and $C$ is north of $B$, the path $A \to B \to C$ turns a square corner at $B$.
- So triangle $ABC$ has a right angle at $B$.
- Marking $D$ another $20$ north of $C$ puts $B$, $C$, $D$ on one vertical line.
💡 East and north always cross at a right angle, so the compass words hand you a right triangle for free.
8.G.B.7 Step 2 Find the eastward leg AB
- In triangle $ABC$ the right angle is at $B$ and $\angle BAC = 45^\circ$, so the third angle is also $45^\circ$ — the triangle is isosceles with $AB = BC$.
- Using the Pythagorean theorem with the two equal legs and hypotenuse $AC = 10\sqrt{2}$: $AB^2 + BC^2 = AC^2$, so $2\,AB^2 = (10\sqrt{2})^2 = 200$, giving $AB^2 = 100$ and $AB = 10$.
- Thus $AB = BC = 10$.
💡 A right triangle with a $45^\circ$ angle has two equal legs, so its two short sides must match.
8.G.B.7 Step 3 Build the big right triangle and get AD
- Now use triangle $ABD$, which also has its right angle at $B$.
- Its horizontal leg is $AB = 10$.
- Its vertical leg is $BD$, and since $B$, $C$, $D$ line up, $BD = BC + CD = 10 + 20 = 30$.
- By the Pythagorean theorem, $AD^2 = AB^2 + BD^2 = 10^2 + 30^2 = 100 + 900 = 1000$, so $AD = \sqrt{1000}$.
💡 $AB$ and the whole north segment $BD$ are the two legs of one big right triangle, so Pythagoras reaches straight to $AD$.
8.NS.A.2 Step 4 Trap the square root between integers
- $AD = \sqrt{1000}$ is not a whole number, so square the candidate integers to bracket it.
- $31^2 = 961$ and $32^2 = 1024$.
- Since $961 < 1000 < 1024$, taking square roots gives $31 < \sqrt{1000} < 32$.
- So $AD$ lies between $31$ and $32$, which is choice (B).
💡 To place a square root between whole numbers, just square those whole numbers and see which pair straddles the inside.
4.G.A.1 Place $A$ at the origin. "East" is the horizontal direction and "north" is the v 8.G.B.7 In triangle $ABC$ the right angle is at $B$ and $\angle BAC = 45^\circ$, so the 8.G.B.7 Now use triangle $ABD$, which also has its right angle at $B$. Its horizontal le 8.NS.A.2 $AD = \sqrt{1000}$ is not a whole number, so square the candidate integers to br Review
Reasonableness: A quick sanity pass: $AD = \sqrt{1000} = 10\sqrt{10} \approx 10 \times 3.162 = 31.6$, which indeed sits between $31$ and $32$. It also should be a touch more than the vertical leg $BD = 30$ (the hypotenuse always beats either leg), and $31.6$ is just over $30$ — consistent. The small eastward leg $AB = 10$ only nudges the length up from $30$ by about $1.6$, so pairs like (D) $33$–$34$ or (E) $34$–$35$ would be far too big.
Alternative: Skip the isosceles reasoning by using coordinates: $A=(0,0)$, $B=(10,0)$ once $AB=10$ is known (or note $AC=10\sqrt2$ at $45^\circ$ means both the east and north components from $A$ to $C$ equal $10$). Then $C=(10,10)$ and $D=(10,30)$, so $AD=\sqrt{10^2+30^2}=\sqrt{1000}$ by the distance formula — the same result.
CCSS standards used (min grade 8)
4.G.A.1Draw points, lines, line segments, rays, angles, and identify in figures (Turning the compass directions east and north into perpendicular lines that create the right angle at $B$.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Finding the legs $AB = BC = 10$ from the $45$-$45$-$90$ triangle and then computing $AD^2 = 10^2 + 30^2 = 1000$.)8.NS.A.2Use rational approximations of irrational numbers to compare their size (Squaring $31$ and $32$ to trap $\sqrt{1000}$ and read off the interval $31 < AD < 32$.)
⭐ East and north make a right angle, so the $45^\circ$ triangle gives $AB = 10$; then one big right triangle gives $AD = \sqrt{10^2+30^2} = \sqrt{1000} \approx 31.6$, which lands between $31$ and $32$.
⭐ East and north make a right angle, so the $45^\circ$ triangle gives $AB = 10$; then one big right triangle gives $AD = \sqrt{10^2+30^2} = \sqrt{1000} \approx 31.6$, which lands between $31$ and $32$.
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