AMC 10 · 2005 · #23
Grade 8 geometry-2dLet AB be a diameter of a circle and C be a point on AB with 2⋅AC=BC. Let D and E be points on the circle such that DC⊥AB and DE is a second diameter. What is the ratio of the area of △DCE to the area of △ABD?
Pick an answer.
AMC 10 2005 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: $\overline{AB}$ is a diameter of a circle and $C$ lies on $\overline{AB}$ with $2\cdot AC = BC$. Points $D$ and $E$ are on the circle with $\overline{DC}\perp\overline{AB}$, and $\overline{DE}$ is a second diameter. Find the ratio of the area of $\triangle DCE$ to the area of $\triangle ABD$.
Givens: $\overline{AB}$ is a diameter of the circle; $C$ is on $\overline{AB}$ with $2\cdot AC = BC$, so $AC:CB = 1:2$; $D$ and $E$ are on the circle, $\overline{DC}\perp\overline{AB}$, and $\overline{DE}$ is a diameter; Answer choices: (A) $\frac{1}{6}$, (B) $\frac{1}{4}$, (C) $\frac{1}{3}$, (D) $\frac{1}{2}$, (E) $\frac{2}{3}$
Unknowns: The ratio $\dfrac{[\triangle DCE]}{[\triangle ABD]}$ of the two triangle areas
Understand
Restated: $\overline{AB}$ is a diameter of a circle and $C$ lies on $\overline{AB}$ with $2\cdot AC = BC$. Points $D$ and $E$ are on the circle with $\overline{DC}\perp\overline{AB}$, and $\overline{DE}$ is a second diameter. Find the ratio of the area of $\triangle DCE$ to the area of $\triangle ABD$.
Givens: $\overline{AB}$ is a diameter of the circle; $C$ is on $\overline{AB}$ with $2\cdot AC = BC$, so $AC:CB = 1:2$; $D$ and $E$ are on the circle, $\overline{DC}\perp\overline{AB}$, and $\overline{DE}$ is a diameter; Answer choices: (A) $\frac{1}{6}$, (B) $\frac{1}{4}$, (C) $\frac{1}{3}$, (D) $\frac{1}{2}$, (E) $\frac{2}{3}$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #4 Introduce a Variable, #7 Identify Subproblems, #3 Eliminate Possibilities
A ratio of areas does not depend on the circle's size, so Tool #4 (Introduce a Variable) lets us pin the radius to $1$ and drop a coordinate grid on the picture. Tool #1 (Draw a Diagram) then turns every named point into a pair of coordinates: once $A$, $B$, $C$, $D$, $E$ are located, both triangles reduce to simple base-times-height calculations. Tool #7 (Identify Subproblems) splits the work into three clean sub-tasks — find $C$ from the ratio, find the height $DC$ from the circle, and find $E$ as the point opposite $D$ — after which each area is one line. Tool #3 (Eliminate Possibilities) matches the exact fraction to the five choices at the end.
Execute — Answer: C
6.RP.A.3 Step 1 Fix the radius and locate C
- The ratio of two areas is unchanged if we scale the whole figure, so choose radius $1$ and diameter $AB=2$.
- Put the center at the origin, so $A=(-1,0)$ and $B=(1,0)$.
- Since $2\cdot AC = BC$, the point $C$ splits $AB$ in the ratio $AC:CB = 1:2$, so $AC$ is one third of $AB$: $AC = \frac{1}{3}\cdot 2 = \frac{2}{3}$.
- Starting at $A=(-1,0)$ and moving $\frac{2}{3}$ toward $B$ gives $C=\left(-\frac{1}{3},\,0\right)$.
💡 A ratio ignores size, so we are free to pick the roundest radius and read positions straight off the axis.
8.G.B.7 Step 2 Find the height DC with the circle
- Because $\overline{DC}\perp\overline{AB}$, point $D$ sits straight above $C$, so $D=\left(-\frac{1}{3},\,y\right)$ for some $y>0$.
- $D$ is on the circle of radius $1$ centered at the origin, so its coordinates satisfy $x^2+y^2=1$.
- Substituting $x=-\frac{1}{3}$ gives $\frac{1}{9}+y^2=1$, hence $y^2=\frac{8}{9}$ and $y=\frac{2\sqrt{2}}{3}$.
- So $D=\left(-\frac{1}{3},\,\frac{2\sqrt{2}}{3}\right)$ and the vertical segment $DC$ has length $\frac{2\sqrt{2}}{3}$.
💡 A point on a circle obeys the Pythagorean relation between its coordinates and the radius.
6.NS.C.6 Step 3 Locate E opposite D
- $\overline{DE}$ is a diameter, so $E$ is the point of the circle directly opposite $D$ through the center.
- Reflecting a point through the origin flips the sign of both coordinates, so $E=\left(\frac{1}{3},\,-\frac{2\sqrt{2}}{3}\right)$.
- Notice $E$'s $x$-coordinate is $\frac{1}{3}$, while $C$ (and $D$) sit at $x=-\frac{1}{3}$: horizontally, $E$ is $\frac{1}{3}-\left(-\frac{1}{3}\right)=\frac{2}{3}$ away from the vertical line through $C$ and $D$.
💡 The two ends of a diameter are mirror images through the center, so one is the other with both signs flipped.
6.G.A.1 Step 4 Area of triangle ABD
- Take $\overline{AB}$ as the base of $\triangle ABD$; it lies along the $x$-axis with length $2$.
- The third vertex is $D$, whose height above the base is its $y$-coordinate, $\frac{2\sqrt{2}}{3}$.
- Using $\text{area}=\frac{1}{2}\cdot\text{base}\cdot\text{height}$ gives $[\triangle ABD]=\frac{1}{2}\cdot 2\cdot\frac{2\sqrt{2}}{3}=\frac{2\sqrt{2}}{3}$.
💡 With the base flat on the axis, the apex's height is simply how high up it sits.
6.G.A.1 Step 5 Area of triangle DCE
- For $\triangle DCE$, use the vertical segment $\overline{DC}$ as the base; it has length $\frac{2\sqrt{2}}{3}$.
- The third vertex is $E$, and its height relative to this vertical base is the horizontal distance from $E$ to the line through $D$ and $C$, which is $\frac{2}{3}$ from Step 3.
- So $[\triangle DCE]=\frac{1}{2}\cdot\frac{2\sqrt{2}}{3}\cdot\frac{2}{3}=\frac{2\sqrt{2}}{9}$.
💡 For a vertical base, the far vertex's 'height' is just how far sideways it reaches.
6.NS.A.1 Step 6 Divide to get the ratio
- Divide the two areas: $\dfrac{[\triangle DCE]}{[\triangle ABD]}=\dfrac{2\sqrt{2}/9}{2\sqrt{2}/3}$.
- The common factor $2\sqrt{2}$ cancels, leaving $\dfrac{1/9}{1/3}=\dfrac{1}{9}\cdot 3=\dfrac{1}{3}$.
- Scanning the choices, $\frac{1}{3}$ is not $\frac{1}{6}$, $\frac{1}{4}$, $\frac{1}{2}$, or $\frac{2}{3}$, so those are ruled out and the ratio matches (C).
- The answer is $\frac{1}{3}$, choice (C).
💡 The messy $2\sqrt{2}$ appears in both areas, so it cancels and only the tidy denominators decide the ratio.
6.RP.A.3 The ratio of two areas is unchanged if we scale the whole figure, so choose radi 8.G.B.7 Because $\overline{DC}\perp\overline{AB}$, point $D$ sits straight above $C$, so 6.NS.C.6 $\overline{DE}$ is a diameter, so $E$ is the point of the circle directly opposi 6.G.A.1 Take $\overline{AB}$ as the base of $\triangle ABD$; it lies along the $x$-axis 6.G.A.1 For $\triangle DCE$, use the vertical segment $\overline{DC}$ as the base; it ha 6.NS.A.1 Divide the two areas: $\dfrac{[\triangle DCE]}{[\triangle ABD]}=\dfrac{2\sqrt{2} Review
Reasonableness: The $2\sqrt{2}$ common factor cancelling is a strong sign the setup is right: a clean ratio like $\frac{1}{3}$ is exactly what a well-posed multiple-choice answer should look like. A second check comes from the shared segment $DC$: both triangles can be seen as having height $DC$, so their ratio is the ratio of the matching bases. For $\triangle ABD$ the base is $AB=2$; for $\triangle DCE$ the relevant width is $E$'s horizontal reach $\frac{2}{3}$, giving $\frac{2/3}{2}=\frac{1}{3}$ — the same answer, so (C) is consistent.
Alternative: Avoid coordinates entirely. Since $\overline{AB}$ is a diameter, $\angle ADB=90^\circ$, so $DC$ is the altitude to the hypotenuse of right triangle $ADB$ and $DC^2=AC\cdot CB$. With $AC=\frac{2}{3}$ and $CB=\frac{4}{3}$ (for diameter $2$), $DC=\frac{2\sqrt2}{3}$. Both $\triangle ABD$ and $\triangle DCE$ share the height $DC$, so the area ratio equals the ratio of their bases measured perpendicular to $DC$: the diameter $AB=2$ for $\triangle ABD$, versus the width $CO+OE$'s horizontal span $\frac{2}{3}$ for $\triangle DCE$. This gives $\frac{2/3}{2}=\frac{1}{3}$, confirming (C) without a single coordinate.
CCSS standards used (min grade 8)
6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Turning $2\cdot AC = BC$ into $AC:CB = 1:2$ to place $C$ one third of the way along $AB$.)8.G.B.7Apply the Pythagorean Theorem to find unknown side lengths (Using $x^2+y^2=1$ for a point on the radius-$1$ circle to find the height $DC=\frac{2\sqrt{2}}{3}$.)6.NS.C.6Understand a rational number as a point on the coordinate plane, including reflections (Finding $E$ as the reflection of $D$ through the origin, flipping the signs of both coordinates.)6.G.A.1Find the area of triangles and other polygons by composing and decomposing (Computing each triangle's area as $\frac{1}{2}\cdot\text{base}\cdot\text{height}$.)6.NS.A.1Interpret and compute quotients of fractions (Dividing the two areas $\frac{2\sqrt{2}}{9}\div\frac{2\sqrt{2}}{3}$ to get the ratio $\frac{1}{3}$.)
⭐ Drop the figure on a grid with radius $1$, read off every point, and both triangles become base-times-height; the ugly $\sqrt{2}$ cancels and the ratio is just $\frac{1}{3}$.
⭐ Drop the figure on a grid with radius $1$, read off every point, and both triangles become base-times-height; the ugly $\sqrt{2}$ cancels and the ratio is just $\frac{1}{3}$.
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