AMC 10 · 2002 · #23
Grade 8 geometry-2dPoints A,B,C and D lie on a line, in that order, with AB=CD and BC=12. Point E is not on the line, and BE=CE=10. The perimeter of △AED is twice the perimeter of △BEC. Find AB.
Pick an answer.
AMC 10 2002 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: Four points $A$, $B$, $C$, $D$ sit on one line in that order, with the two outer gaps equal ($AB = CD$) and the middle gap $BC = 12$. A point $E$ off the line is the same distance from $B$ and from $C$: $BE = CE = 10$. The perimeter of the big triangle $AED$ is exactly twice the perimeter of the small triangle $BEC$. Find the length $AB$.
Givens: $A$, $B$, $C$, $D$ are collinear in that order; $AB = CD$ (the two outer segments are equal); $BC = 12$; $E$ is off the line with $BE = CE = 10$; perimeter of $\triangle AED$ = 2 $\times$ perimeter of $\triangle BEC$; Answer choices: (A) $15/2$, (B) $8$, (C) $17/2$, (D) $9$, (E) $19/2$
Unknowns: the length $AB$
Understand
Restated: Four points $A$, $B$, $C$, $D$ sit on one line in that order, with the two outer gaps equal ($AB = CD$) and the middle gap $BC = 12$. A point $E$ off the line is the same distance from $B$ and from $C$: $BE = CE = 10$. The perimeter of the big triangle $AED$ is exactly twice the perimeter of the small triangle $BEC$. Find the length $AB$.
Givens: $A$, $B$, $C$, $D$ are collinear in that order; $AB = CD$ (the two outer segments are equal); $BC = 12$; $E$ is off the line with $BE = CE = 10$; perimeter of $\triangle AED$ = 2 $\times$ perimeter of $\triangle BEC$; Answer choices: (A) $15/2$, (B) $8$, (C) $17/2$, (D) $9$, (E) $19/2$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #4 Introduce a Variable, #7 Identify Subproblems, #13 Convert to Algebra
A clear picture is the whole game here. Once you plot the four points and $E$, the equal distances $BE = CE$ (and $AB = CD$) make the figure mirror-symmetric, so $E$ sits directly above the midpoint of $BC$ — and of $AD$. Dropping that vertical altitude (Tool #1) splits each triangle into right triangles, which is exactly what the Pythagorean theorem needs. The perimeter condition can't be used until we name the unknown, so we let $AB = x$ (Tool #4) and write both $AD$ and the slant side $AE$ in terms of $x$. That gives two expressions for the same length $AE$; setting them equal (Tool #13) collapses to one equation for $x$. Tool #7 (Identify Subproblems) keeps the two jobs — find the altitude, then find $AE$ — separate and clean.
Execute — Answer: D
8.G.B.7 Step 1 Draw it and drop the altitude
- Plot $A$, $B$, $C$, $D$ left to right, with $E$ above the line.
- Since $BE = CE = 10$, triangle $BEC$ is isosceles, so $E$ sits directly above the midpoint $M$ of $BC$ and the segment $EM$ is perpendicular to the line.
- That makes $BM = MC = \tfrac{12}{2} = 6$.
- In right triangle $BME$ the Pythagorean theorem gives the height $EM = \sqrt{BE^2 - BM^2} = \sqrt{10^2 - 6^2} = \sqrt{64} = 8$.
💡 An isosceles triangle's tip sits straight above the middle of its base, so the altitude splits the base into two equal halves you can measure.
6.EE.B.6 Step 2 Name AB and use the perimeter fact
- The small triangle has perimeter $BE + EC + BC = 10 + 10 + 12 = 32$, so the big triangle $AED$ must have perimeter $2 \times 32 = 64$.
- Let $AB = x$; then $CD = x$ too, and $AD = AB + BC + CD = x + 12 + x = 2x + 12$.
- Because the figure is symmetric about the vertical line $EM$, the two slant sides are equal: $AE = DE$.
- So the perimeter of $\triangle AED$ is $AD + 2\,AE = 64$, which rearranges to $AE = \dfrac{64 - (2x + 12)}{2} = 26 - x$.
💡 Giving the unknown a letter turns the words "twice the perimeter" into an equation you can actually push around.
8.G.B.7 Step 3 Measure AE from the right triangle
- Because $AB = CD$, the midpoint $M$ of $BC$ is also the midpoint of the whole segment $AD$, so the same altitude $EM = 8$ drops to $AD$.
- The horizontal distance from $A$ to the foot $M$ is $AM = AB + BM = x + 6$.
- Triangle $AME$ is right-angled at $M$, so the Pythagorean theorem gives a second expression for $AE$: $AE^2 = AM^2 + EM^2 = (x + 6)^2 + 8^2$.
💡 The vertical altitude turns the slanted side $AE$ into the hypotenuse of a right triangle whose legs you already know.
8.EE.C.7 Step 4 Set the two AE's equal and solve
- Now $AE$ is written two ways: $AE = 26 - x$ from the perimeter, and $AE^2 = (x+6)^2 + 64$ from the right triangle.
- Square the first and match it to the second: $(26 - x)^2 = (x + 6)^2 + 64$.
- Expanding both sides gives $676 - 52x + x^2 = x^2 + 12x + 36 + 64$.
- The $x^2$ terms cancel, leaving $676 - 52x = 12x + 100$, so $576 = 64x$ and $x = 9$.
- Therefore $AB = 9$, which is choice (D).
💡 Two honest descriptions of the same length must agree, and forcing them to agree pins down the one unknown.
8.G.B.7 Plot $A$, $B$, $C$, $D$ left to right, with $E$ above the line. Since $BE = CE = 6.EE.B.6 The small triangle has perimeter $BE + EC + BC = 10 + 10 + 12 = 32$, so the big 8.G.B.7 Because $AB = CD$, the midpoint $M$ of $BC$ is also the midpoint of the whole se 8.EE.C.7 Now $AE$ is written two ways: $AE = 26 - x$ from the perimeter, and $AE^2 = (x+6 Review
Reasonableness: Plug $x = 9$ back in. Then $AM = 9 + 6 = 15$ and $EM = 8$, so $AE = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17$ — a clean $8\text{-}15\text{-}17$ right triangle, and it matches $26 - x = 26 - 9 = 17$. The big triangle then has $AD = 2(9) + 12 = 30$ and two slant sides of $17$, giving perimeter $30 + 17 + 17 = 64$, which is exactly twice the small triangle's $32$. Everything lines up, so (D) is correct. The half-integer choices like $15/2$ or $17/2$ are traps for anyone who mishandles the factor of two.
Alternative: Put coordinates on the picture: $M = (0,0)$, $E = (0,8)$, $B = (-6,0)$, $C = (6,0)$, and by symmetry $A = (-(6+x),0)$, $D = (6+x,0)$. Then $AE = DE = \sqrt{(6+x)^2 + 64}$, and the perimeter equation $2\sqrt{(6+x)^2+64} + (2x+12) = 64$ leads to the same $\sqrt{(6+x)^2+64} = 26 - x$ and hence $x = 9$. Same answer, reached by computing the distances directly instead of reasoning about symmetry.
CCSS standards used (min grade 8)
8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Finding the altitude $EM = \sqrt{10^2 - 6^2} = 8$ and expressing the slant side as $AE^2 = (x+6)^2 + 64$.)6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Naming $AB = x$ and writing $AD = 2x + 12$ and $AE = 26 - x$ from the perimeter condition.)8.EE.C.7Solve linear equations in one variable (Solving $676 - 52x = 12x + 100$ (after the $x^2$ terms cancel) to get $x = 9$.)
⭐ Drop the altitude to split the figure into right triangles, write the slant side two ways — once from the doubled perimeter and once from the Pythagorean theorem — and make them agree to pin down $AB = 9$.
⭐ Drop the altitude to split the figure into right triangles, write the slant side two ways — once from the doubled perimeter and once from the Pythagorean theorem — and make them agree to pin down $AB = 9$.
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