AMC 10 · 2012 · #21
Grade 8 geometry-2dFour distinct points are arranged on a plane so that the segments connecting them have lengths a, a, a, a, 2a, and b. What is the ratio of b to a?
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: Place four distinct points in a plane. Between them there are six connecting segments, and their lengths are $a$, $a$, $a$, $a$, $2a$, and $b$ in some order. Figure out how the four points must be arranged, then find the value of the ratio $b:a$.
Givens: There are $4$ distinct points, so there are $\binom{4}{2}=6$ segments between them; Four of the six lengths equal $a$; One of the six lengths equals $2a$; The last length is $b$; Answer choices: (A) $\sqrt{3}$, (B) $2$, (C) $\sqrt{5}$, (D) $3$, (E) $\pi$
Unknowns: The arrangement of the four points that makes these six lengths possible; The ratio $b$ to $a$
Understand
Restated: Place four distinct points in a plane. Between them there are six connecting segments, and their lengths are $a$, $a$, $a$, $a$, $2a$, and $b$ in some order. Figure out how the four points must be arranged, then find the value of the ratio $b:a$.
Givens: There are $4$ distinct points, so there are $\binom{4}{2}=6$ segments between them; Four of the six lengths equal $a$; One of the six lengths equals $2a$; The last length is $b$; Answer choices: (A) $\sqrt{3}$, (B) $2$, (C) $\sqrt{5}$, (D) $3$, (E) $\pi$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #3 Eliminate Possibilities, #7 Identify Subproblems, #8 Analyze the Units
The problem is entirely about positions and distances, so Tool #1 (Draw a Diagram) is the natural driver: the answer falls out once the four points are pinned down on paper. The single long segment $2a$ is the pressure point. Tool #3 (Eliminate Possibilities) uses the triangle inequality to rule out any real triangle with sides $a,a,2a$, which forces three of the points onto one straight line. Tool #7 (Identify Subproblems) then splits the six segments into two clean pieces — the three collinear segments $a,a,2a$ and the three segments from the fourth point. Tool #8 (Analyze the Units) keeps every length written as a multiple of $a$, so the final answer comes out as a pure ratio with the $a$ dividing away.
Execute — Answer: A
7.G.A.2 Step 1 Zoom in on the long segment
- Six segments join the four points.
- Give the longest one, $2a$, its own attention.
- Call its two endpoints $P$ and $Q$, so $PQ = 2a$.
- Ask what happens if some third point $R$ is joined to both $P$ and $Q$ by short $a$-segments.
- That would make a triangle $PQR$ with sides $a$, $a$, and $2a$.
💡 The one length that is twice as long as all the others is what controls the whole shape, so start there.
7.G.A.2 Step 2 Rule out the flat triangle
- A real triangle must have each side shorter than the sum of the other two.
- For sides $a$, $a$, $2a$ we get $a + a = 2a$, which is not greater than $2a$.
- So $P$, $R$, $Q$ cannot make a real triangle: they must be flattened onto one straight line, with $R$ exactly in the middle of segment $PQ$.
- That is the only way a point can sit at distance $a$ from both ends of a segment of length $2a$.
💡 When two sides add up to exactly the third, the triangle collapses flat onto a line.
5.G.A.2 Step 3 Put the three collinear points on a number line
- Lay the line down as the $x$-axis.
- Place $P=(0,0)$, the midpoint $R=(a,0)$, and $Q=(2a,0)$.
- This uses three of the six segments: $PR=a$, $RQ=a$, and $PQ=2a$.
- The remaining three segments all reach out to the fourth point, which we call $D$, and they must be the leftover lengths $a$, $a$, and $b$.
💡 Coordinates turn distance questions into arithmetic you can just compute.
7.G.A.2 Step 4 Locate the fourth point
- Point $D$ must supply the lengths $a$, $a$, $b$ to $P$, $R$, $Q$.
- Two of those three distances equal $a$.
- They cannot be $DP=a$ and $DQ=a$, because the only point at distance $a$ from both ends of the $2a$ segment is the midpoint $R$, and $D$ has to be a new point.
- So the two $a$-distances go to two neighbors that are only $a$ apart: take $DP=a$ and $DR=a$.
- Then $D$ caps an equilateral triangle on $PR$, sitting at $D=\left(\tfrac{a}{2},\tfrac{\sqrt{3}}{2}a\right)$.
💡 Two equal $a$-legs on a base of length $a$ build the familiar equilateral triangle.
8.G.B.8 Step 5 Measure the last segment
- The one segment not yet named is $DQ$, and its length is $b$.
- Use the distance between $D=\left(\tfrac{a}{2},\tfrac{\sqrt{3}}{2}a\right)$ and $Q=(2a,0)$.
- The horizontal gap is $2a-\tfrac{a}{2}=\tfrac{3a}{2}$ and the vertical gap is $\tfrac{\sqrt{3}}{2}a$.
- The Pythagorean theorem gives $b^2 = \left(\tfrac{3a}{2}\right)^2 + \left(\tfrac{\sqrt{3}}{2}a\right)^2 = \tfrac{9a^2}{4}+\tfrac{3a^2}{4}=3a^2$, so $b=\sqrt{3}\,a$.
💡 The distance between two plotted points is just the Pythagorean theorem on the horizontal and vertical gaps.
6.RP.A.1 Step 6 Form the ratio
- Now every length checks out: the four $a$-segments are $PR$, $RQ$, $DP$, $DR$; the long one is $PQ=2a$; and the last is $DQ=b=\sqrt{3}\,a$.
- The ratio of $b$ to $a$ is $\dfrac{\sqrt{3}\,a}{a}=\sqrt{3}$.
- So the answer is $(\text{A})\ \sqrt{3}$.
💡 Writing both lengths as multiples of $a$ lets the $a$ cancel, leaving a pure number for the ratio.
7.G.A.2 Six segments join the four points. Give the longest one, $2a$, its own attention 7.G.A.2 A real triangle must have each side shorter than the sum of the other two. For s 5.G.A.2 Lay the line down as the $x$-axis. Place $P=(0,0)$, the midpoint $R=(a,0)$, and 7.G.A.2 Point $D$ must supply the lengths $a$, $a$, $b$ to $P$, $R$, $Q$. Two of those t 8.G.B.8 The one segment not yet named is $DQ$, and its length is $b$. Use the distance b 6.RP.A.1 Now every length checks out: the four $a$-segments are $PR$, $RQ$, $DP$, $DR$; t Review
Reasonableness: Count the segments in the final picture: $PR,\,RQ,\,DP,\,DR$ are each $a$ (four of them), $PQ=2a$ is the long one, and $DQ=\sqrt{3}\,a=b$. That is exactly the required list $a,a,a,a,2a,b$, and all four points are distinct, so the arrangement is valid. The value $\sqrt{3}\approx 1.73$ is sensibly between $a$ and $2a$, which fits a segment that stretches from the equilateral peak across to the far end of the base line. Among the choices only $\sqrt{3}$ matches, confirming (A).
Alternative: Skip coordinates and use the isosceles triangle $DPQ$ directly. Point $D$ sits at distance $a$ from midpoint $R$, and $DR$ is the median to side $PQ$ of the triangle $DPQ$ where $DP=a$ and $PQ=2a$ with $R$ the midpoint. Because $D$, $P$, $R$ form an equilateral triangle, $DR \perp PQ$ is not assumed — instead drop the height and use the median-length relation, or simply note $\triangle DRQ$ has legs $DR=a$, $RQ=a$ meeting at the $120^\circ$ angle left over next to the $60^\circ$ equilateral corner. The Law of Cosines then gives $DQ^2 = a^2 + a^2 - 2a^2\cos 120^\circ = 2a^2 + a^2 = 3a^2$, so $DQ=\sqrt{3}\,a$ and $b/a=\sqrt{3}$ — the same answer.
CCSS standards used (min grade 8)
7.G.A.2Draw geometric shapes with given conditions including triangles (Using the triangle inequality to reject a real triangle with sides $a,a,2a$ and to build the equilateral triangle that fixes the fourth point.)5.G.A.2Represent real-world and mathematical problems by graphing points (Placing the three collinear points at $(0,0)$, $(a,0)$, and $(2a,0)$ so distances become computable.)8.G.B.8Apply the Pythagorean theorem to find distance between two points in a coordinate system (Computing the last segment $b=DQ=\sqrt{3}\,a$ from the horizontal and vertical gaps between $D$ and $Q$.)6.RP.A.1Understand the concept of a ratio and use ratio language (Forming the ratio $b:a$ and simplifying $\tfrac{\sqrt{3}\,a}{a}=\sqrt{3}$.)
⭐ The length $2a$ is too long to be a triangle side with two $a$-sides, so three points must lie flat on a line; the fourth point makes an equilateral triangle, and the last segment measures $\sqrt{3}\,a$, giving $b/a=\sqrt{3}$.
⭐ The length $2a$ is too long to be a triangle side with two $a$-sides, so three points must lie flat on a line; the fourth point makes an equilateral triangle, and the last segment measures $\sqrt{3}\,a$, giving $b/a=\sqrt{3}$.
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