AMC 10 · 2009 · #24
Grade 8 geometry-2dThe keystone arch is an ancient architectural feature. It is composed of congruent isosceles trapezoids fitted together along the non-parallel sides, as shown. The bottom sides of the two end trapezoids are horizontal. In an arch made with 9 trapezoids, let x be the angle measure in degrees of the larger interior angle of the trapezoid. What is x?
Pick an answer.
AMC 10 2009 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: A keystone arch is built from $9$ congruent isosceles trapezoids joined leg-to-leg. Together they curve from the ground on one side, over the top, back down to the ground on the other side. The bottom sides of the two end trapezoids rest flat on the ground. Each trapezoid has two equal (larger) interior angles and two equal (smaller) interior angles. Find the measure $x$, in degrees, of the larger interior angle.
Givens: The arch is made of $9$ congruent isosceles trapezoids fitted together along their non-parallel sides (their legs).; The bottom side of each of the two end trapezoids is horizontal, so the arch rises from flat ground on both ends.; In an isosceles trapezoid the two angles along the longer parallel side are equal (the smaller pair), and the two angles along the shorter parallel side are equal (the larger pair).; Answer choices: (A) $100$, (B) $102$, (C) $104$, (D) $106$, (E) $108$.
Unknowns: $x$, the degree measure of the larger interior angle of one trapezoid.
Understand
Restated: A keystone arch is built from $9$ congruent isosceles trapezoids joined leg-to-leg. Together they curve from the ground on one side, over the top, back down to the ground on the other side. The bottom sides of the two end trapezoids rest flat on the ground. Each trapezoid has two equal (larger) interior angles and two equal (smaller) interior angles. Find the measure $x$, in degrees, of the larger interior angle.
Givens: The arch is made of $9$ congruent isosceles trapezoids fitted together along their non-parallel sides (their legs).; The bottom side of each of the two end trapezoids is horizontal, so the arch rises from flat ground on both ends.; In an isosceles trapezoid the two angles along the longer parallel side are equal (the smaller pair), and the two angles along the shorter parallel side are equal (the larger pair).; Answer choices: (A) $100$, (B) $102$, (C) $104$, (D) $106$, (E) $108$.
Plan
Primary tool: #1 Draw a Diagram
Secondary: #17 Visualize Spatial Relationships, #7 Identify Subproblems
The picture hides a clean triangle. If you extend the two legs of every trapezoid (Tool #1, an auxiliary construction), symmetry forces all of them to meet at one shared point $O$ in the middle of the base line, and each trapezoid becomes a thin triangle with its tip cut off at $O$. Tool #17 (Visualize Spatial Relationships) sees that the $9$ tip-triangles fan out to fill the half-turn above the base, so each tip angle is a fair share of $180^\circ$. Tool #7 (Identify Subproblems) then chains three easy facts: tip angle $\to$ base angles of the isosceles tip-triangle $\to$ the trapezoid's larger angle by the same-side-interior rule.
Execute — Answer: A
4.G.A.1 Step 1 Extend the legs to a shared point
- Take one trapezoid and draw its two slanted legs outward until they cross.
- Do this for every trapezoid.
- Because neighboring trapezoids are mirror images across their shared leg, each pair of outside legs meets that shared leg's extension at the same place.
- Repeating this across all $9$ pieces, every leg extension passes through one common point $O$, and $O$ sits in the middle of the base line.
- Each trapezoid is now the top slice of a triangle whose tip is at $O$.
💡 Sliding the slanted sides until they meet turns each trapezoid into an easy-to-measure triangle.
4.MD.C.7 Step 2 Share the half-turn among 9 tips
- The arch starts flat on the ground, rises over the top, and comes back down flat on the other side.
- Going from the flat left end to the flat right end is a half-turn, which is $180^\circ$.
- All $9$ tip angles at $O$ sit side by side and together fill that half-turn.
- Since the trapezoids are congruent, the tip angles are all equal, so each one is $180^\circ$ divided by $9$.
💡 The nine equal wedges pave the same half-turn a straight road would make, so each wedge is one-ninth of it.
8.G.A.5 Step 3 Find the trapezoid's smaller angle
- Look at one tip-triangle at $O$.
- Its two long sides are the extended legs, and they are equal in length, so the triangle is isosceles.
- Its three angles add to $180^\circ$.
- The tip angle is $20^\circ$, so the two equal base angles share the remaining $160^\circ$, giving $80^\circ$ each.
- The trapezoid's bottom edge lies along the base of this triangle, so the trapezoid's angle at that longer parallel side equals a base angle: the smaller interior angle of the trapezoid is $80^\circ$.
💡 In a triangle the three angles total $180^\circ$, so once the tip is known the two equal base angles are forced.
7.G.B.5 Step 4 Turn the smaller angle into the larger angle
- Inside a trapezoid, one leg meets the two parallel sides.
- The smaller angle ($80^\circ$) and the larger angle $x$ sit on the same leg, one at each parallel side, so they are same-side interior angles between the parallels and add to $180^\circ$.
- Therefore $x = 180^\circ - 80^\circ = 100^\circ$.
- So the larger interior angle is $100^\circ$, which is choice (A).
💡 Along one leg the trapezoid's two angles fill a straight line between the parallel sides, so they must sum to $180^\circ$.
4.G.A.1 Take one trapezoid and draw its two slanted legs outward until they cross. Do th 4.MD.C.7 The arch starts flat on the ground, rises over the top, and comes back down flat 8.G.A.5 Look at one tip-triangle at $O$. Its two long sides are the extended legs, and t 7.G.B.5 Inside a trapezoid, one leg meets the two parallel sides. The smaller angle ($80 Review
Reasonableness: A quick sanity check: the smaller and larger angles of an isosceles trapezoid must add to $180^\circ$ along each leg, and $80^\circ + 100^\circ = 180^\circ$ works. The larger angle should be a bit more than a right angle because the pieces lean only slightly, and $100^\circ$ is just past $90^\circ$, which matches a gently curving arch. If each larger angle were as big as $108^\circ$ the smaller angle would be $72^\circ$, forcing a tip angle of $180^\circ - 2(72^\circ) = 36^\circ$ and only $180^\circ / 36^\circ = 5$ pieces per half circle, not $9$. Choice (A) is the only one that fits $9$ trapezoids.
Alternative: Instead of the tip-triangle, build a polygon. Reflect the arch across the base line to complete a full ring, or close off the arch with the base to form a big polygon, and use the polygon interior-angle sum. Setting the total of all the trapezoid angles plus the base angles equal to that sum gives the same equation and again yields the smaller angle $80^\circ$ and the larger angle $x = 100^\circ$.
CCSS standards used (min grade 8)
4.G.A.1Draw points, lines, line segments, rays, angles, and identify in figures (Drawing the auxiliary construction: extending each trapezoid's legs until they meet at the shared point $O$.)4.MD.C.7Recognize angle measure as additive and solve addition and subtraction problems (Treating the $9$ equal tip angles at $O$ as adding up to the $180^\circ$ half-turn, giving each tip angle $= 180^\circ/9 = 20^\circ$.)8.G.A.5Use informal arguments to establish facts about angle sum and exterior angles (Using the triangle angle sum on the isosceles tip-triangle to get the two equal base angles $(180^\circ-20^\circ)/2 = 80^\circ$.)7.G.B.5Use facts about supplementary and adjacent angles to solve for an unknown angle (Using the same-side interior angles along one leg ($80^\circ + x = 180^\circ$) to find the larger angle $x = 100^\circ$.)
⭐ Slide the slanted sides until they meet at one center point, share the half-turn of $180^\circ$ among the nine pieces, and the rest is just triangle and straight-line angles.
⭐ Slide the slanted sides until they meet at one center point, share the half-turn of $180^\circ$ among the nine pieces, and the rest is just triangle and straight-line angles.
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