AMC 10 · 2004 · #20
Grade 7 geometry-2dIn △ABC points D and E lie on BC and AC, respectively. If AD and BE intersect at T so that DTAT=3 and ETBT=4, what is BDCD?
Pick an answer.
AMC 10 2004 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: In triangle $ABC$, point $D$ is on side $BC$ and point $E$ is on side $AC$. The cevians $AD$ and $BE$ cross at $T$. Along $AD$, the piece $AT$ is $3$ times the piece $DT$; along $BE$, the piece $BT$ is $4$ times the piece $ET$. Find how the point $D$ splits $BC$, that is, the ratio $\frac{CD}{BD}$.
Givens: $D$ lies on $BC$ and $E$ lies on $AC$.; $AD$ and $BE$ meet at the interior point $T$.; $\frac{AT}{DT}=3$, so $T$ divides $AD$ with $AT:TD=3:1$.; $\frac{BT}{ET}=4$, so $T$ divides $BE$ with $BT:TE=4:1$.; Answer choices: (A) $\frac{1}{8}$, (B) $\frac{2}{9}$, (C) $\frac{3}{10}$, (D) $\frac{4}{11}$, (E) $\frac{5}{12}$.
Unknowns: The ratio $\frac{CD}{BD}$ in which $D$ divides side $BC$.
Understand
Restated: In triangle $ABC$, point $D$ is on side $BC$ and point $E$ is on side $AC$. The cevians $AD$ and $BE$ cross at $T$. Along $AD$, the piece $AT$ is $3$ times the piece $DT$; along $BE$, the piece $BT$ is $4$ times the piece $ET$. Find how the point $D$ splits $BC$, that is, the ratio $\frac{CD}{BD}$.
Givens: $D$ lies on $BC$ and $E$ lies on $AC$.; $AD$ and $BE$ meet at the interior point $T$.; $\frac{AT}{DT}=3$, so $T$ divides $AD$ with $AT:TD=3:1$.; $\frac{BT}{ET}=4$, so $T$ divides $BE$ with $BT:TE=4:1$.; Answer choices: (A) $\frac{1}{8}$, (B) $\frac{2}{9}$, (C) $\frac{3}{10}$, (D) $\frac{4}{11}$, (E) $\frac{5}{12}$.
Plan
Primary tool: #4 Introduce a Variable
Secondary: #7 Identify Subproblems, #1 Draw a Diagram
The picture has no measurable lengths — only two ratios along the two cevians — so I trade lengths for areas, which the ratios control directly. The key fact (Tool #7, Identify Subproblems) is that two triangles sharing the same height have areas in the ratio of their bases. So I cut triangle $ABC$ into the four small triangles meeting at $T$ and let one of them be my unit. Then I name the one unknown area with a variable (Tool #4, Introduce a Variable): the area of triangle $TCD$. Reading triangle $ACT$'s area two different ways gives a single equation in that variable, and that variable turns out to equal the answer $\frac{CD}{BD}$ itself. Tool #1 (Draw a Diagram) keeps the four pieces and which cevian controls which ratio straight.
Execute — Answer: D
6.G.A.1 Step 1 Turn the length ratio into an area ratio
- Two triangles that share the same height have areas in the ratio of their bases, because area $=\tfrac12\cdot\text{base}\cdot\text{height}$.
- Triangles $TCD$ and $TBD$ share the apex $T$, and their bases $CD$ and $BD$ lie on the same line $BC$, so they have the same height from $T$.
- Therefore $\frac{CD}{BD}=\frac{[TCD]}{[TBD]}$, where $[\,\cdot\,]$ means area.
- This converts the whole problem into finding a ratio of two of the small triangles around $T$.
💡 Same height means the bigger base makes the bigger triangle, in exact proportion.
7.RP.A.2 Step 2 Use the AD ratio: set a unit and a variable
- Let $[TBD]=1$ be my unit of area, and let $[TCD]=x$ be the unknown; by Step 1 the answer is exactly $x$.
- Now use $AT:TD=3:1$.
- Triangles $ABT$ and $DBT$ share apex $B$ with bases $AT$ and $TD$ on line $AD$, so $[ABT]=3\,[DBT]=3$.
- Triangles $ACT$ and $DCT$ share apex $C$ with the same bases $AT,TD$, so $[ACT]=3\,[DCT]=3x$.
💡 Because $AT$ is three times $TD$, every triangle built on $AT$ is three times its partner built on $TD$.
7.RP.A.2 Step 3 Use the BE ratio on the same pieces
- Now use $BT:TE=4:1$.
- Triangles $ABT$ and $AET$ share apex $A$ with bases $BT$ and $ET$ on line $BE$, so $[AET]=\tfrac14[ABT]=\tfrac34$.
- Triangles $CBT$ and $CET$ share apex $C$ with the same bases $BT,ET$, so $[CET]=\tfrac14[CBT]$.
- The triangle $CBT$ is itself split by $D$ into $TBD$ and $TCD$, so $[CBT]=[TBD]+[TCD]=1+x$, giving $[CET]=\frac{1+x}{4}$.
💡 The same $4:1$ split of $BE$ shrinks each triangle on $ET$ to a quarter of its partner on $BT$.
7.EE.B.4 Step 4 Read triangle ACT two ways and solve
- Point $E$ is on $AC$, so segment $TE$ cuts triangle $ACT$ into $AET$ and $CET$: thus $[ACT]=[AET]+[CET]=\tfrac34+\frac{1+x}{4}$.
- But Step 2 already found $[ACT]=3x$.
- Setting the two expressions equal gives $3x=\tfrac34+\frac{1+x}{4}$.
- Multiply through by $4$: $12x=3+(1+x)=4+x$, so $11x=4$ and $x=\frac{4}{11}$.
- Since the answer is $\frac{CD}{BD}=x$, we get $\frac{CD}{BD}=\frac{4}{11}$, which is choice (D).
💡 One area written two different ways must agree, and that single equation pins down the unknown.
6.G.A.1 Two triangles that share the same height have areas in the ratio of their bases, 7.RP.A.2 Let $[TBD]=1$ be my unit of area, and let $[TCD]=x$ be the unknown; by Step 1 th 7.RP.A.2 Now use $BT:TE=4:1$. Triangles $ABT$ and $AET$ share apex $A$ with bases $BT$ an 7.EE.B.4 Point $E$ is on $AC$, so segment $TE$ cuts triangle $ACT$ into $AET$ and $CET$: Review
Reasonableness: The answer is a ratio between $0$ and $1$, which fits: $D$ sits partway along $BC$, and the strong pull of the cevians ($AT$ three times $TD$, $BT$ four times $ET$) should keep $D$ closer to $B$ than to $C$, so $CD>BD$ would be wrong — indeed $\frac{4}{11}<1$ means $CD<BD$, consistent with $D$ near $B$. All areas came out positive ($[TBD]=1,[TCD]=\frac4{11},[ABT]=3,[AET]=\frac34$), so no piece was forced negative, and $\frac{4}{11}$ is exactly one of the offered choices.
Alternative: Mass points give the same answer fast. Put mass $1$ at $A$; since $AT:TD=3:1$ needs $D$ to carry mass $3$, and $D$ balances $B$ and $C$, we need $m_B+m_C=3$. Since $BT:TE=4:1$ needs $E$ to carry mass $4m_B$ balanced against $B$, and $E$ balances $A$ and $C$, we need $m_A+m_C=4m_B$, i.e. $1+m_C=4m_B$. With $m_C=3-m_B$ this gives $4=5m_B$, so $m_B=\frac45$, $m_C=\frac{11}5$. Then $\frac{CD}{BD}=\frac{m_B}{m_C}=\frac{4/5}{11/5}=\frac{4}{11}$.
CCSS standards used (min grade 7)
6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Using area $=\tfrac12\cdot$base$\cdot$height to see that triangles with a shared height have areas in the ratio of their bases, and composing/decomposing $ABC$ into the small triangles around $T$.)7.RP.A.2Recognize and represent proportional relationships between quantities (Turning $AT:TD=3:1$ and $BT:TE=4:1$ into proportional area statements such as $[ABT]=3[DBT]$ and $[CET]=\tfrac14[CBT]$.)7.EE.B.4Use variables to represent quantities and construct simple equations and inequalities (Naming the unknown area $x$, writing $[ACT]$ two ways as the equation $3x=\tfrac34+\frac{1+x}{4}$, and solving $11x=4$.)
⭐ Trade lengths for areas: triangles with the same height compare by their bases, so name one small triangle's area $x$, write another triangle's area two different ways, and the single equation hands you $\frac{CD}{BD}=\frac{4}{11}$.
⭐ Trade lengths for areas: triangles with the same height compare by their bases, so name one small triangle's area $x$, write another triangle's area two different ways, and the single equation hands you $\frac{CD}{BD}=\frac{4}{11}$.
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