AMC 10 · 2013 · #23
Grade 8 geometry-2dIn triangle ABC, AB=13, BC=14, and CA=15. Distinct points D, E, and F lie on segments BC, CA, and DE, respectively, such that AD⊥BC, DE⊥AC, and AF⊥BF. The length of segment DF can be written as nm, where m and n are relatively prime positive integers. What is m+n?
Pick an answer.
AMC 10 2013 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 $ABC$ with $AB=13$, $BC=14$, $CA=15$, point $D$ is the foot of the altitude from $A$ to $BC$, point $E$ is the foot of the perpendicular from $D$ to $AC$, and $F$ lies on segment $\overline{DE}$ with $\overline{AF}\perp\overline{BF}$. Write $DF=\frac{m}{n}$ in lowest terms and find $m+n$.
Givens: $AB=13$, $BC=14$, $CA=15$ (the classic $13$-$14$-$15$ triangle); $D$ on $\overline{BC}$ with $\overline{AD}\perp\overline{BC}$ (altitude foot); $E$ on $\overline{CA}$ with $\overline{DE}\perp\overline{AC}$; $F$ on $\overline{DE}$ with $\overline{AF}\perp\overline{BF}$; Answer choices: (A) $18$, (B) $21$, (C) $24$, (D) $27$, (E) $30$
Unknowns: The length $DF=\frac{m}{n}$, and then $m+n$
Understand
Restated: In triangle $ABC$ with $AB=13$, $BC=14$, $CA=15$, point $D$ is the foot of the altitude from $A$ to $BC$, point $E$ is the foot of the perpendicular from $D$ to $AC$, and $F$ lies on segment $\overline{DE}$ with $\overline{AF}\perp\overline{BF}$. Write $DF=\frac{m}{n}$ in lowest terms and find $m+n$.
Givens: $AB=13$, $BC=14$, $CA=15$ (the classic $13$-$14$-$15$ triangle); $D$ on $\overline{BC}$ with $\overline{AD}\perp\overline{BC}$ (altitude foot); $E$ on $\overline{CA}$ with $\overline{DE}\perp\overline{AC}$; $F$ on $\overline{DE}$ with $\overline{AF}\perp\overline{BF}$; Answer choices: (A) $18$, (B) $21$, (C) $24$, (D) $27$, (E) $30$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #4 Introduce a Variable, #7 Identify Subproblems, #13 Convert to Algebra
Three stacked perpendiculars scream 'set up coordinates.' Tool #1 (Diagram): sketch the $13$-$14$-$15$ triangle, drop the altitude $AD$, and pin the figure to axes so every 'perpendicular' becomes a slope condition. Tool #7 (Subproblems): first pin down $D$ and $A$ from the altitude, then find the line $DE$, then locate $F$ on it. Tool #4 (Introduce a Variable): let $t=DF$ and ride the line $DE$ a distance $t$ from $D$. Tool #13 (Convert to Algebra): the right angle at $F$ (that $\overline{AF}\perp\overline{BF}$) turns into one equation in $t$.
Execute — Answer: B
8.G.B.7 Step 1 Split the base with the altitude
- The $13$-$14$-$15$ triangle splits neatly along its altitude to $BC$.
- If $AD=h$ and $BD=x$, then $x^2+h^2=13^2$ and $(14-x)^2+h^2=15^2$.
- Subtracting gives $28x-196=169-225$, so $x=5$; then $h=\sqrt{169-25}=12$.
- So $BD=5$, $DC=9$, and $AD=12$.
- (You may also recognize the $5$-$12$-$13$ and $9$-$12$-$15$ right triangles sharing the leg $12$.)
💡 The altitude chops the big triangle into two right triangles that share the same height, so Pythagoras pins down every length.
5.G.A.2 Step 2 Put the figure on axes
- Place $D$ at the origin with $BC$ along the $x$-axis.
- Since $BD=5$ and $DC=9$, put $B=(-5,0)$ and $C=(9,0)$.
- Because $\overline{AD}\perp\overline{BC}$ and $AD=12$, put $A=(0,12)$.
- This choice makes the altitude the $y$-axis, so the perpendicularities become clean slope statements.
💡 Anchoring the right angle at the origin lets each 'perpendicular' in the problem show up as a slope you can read off.
8.EE.B.6 Step 3 Find the direction of line DE
- Line $AC$ runs from $A=(0,12)$ to $C=(9,0)$, so its slope is $\frac{0-12}{9-0}=-\frac{4}{3}$.
- Since $\overline{DE}\perp\overline{AC}$, line $DE$ has the negative-reciprocal slope $\frac{3}{4}$, and it passes through $D=(0,0)$: so $DE$ is the line $y=\frac{3}{4}x$.
- A convenient unit step along it is $\left(\tfrac{4}{5},\tfrac{3}{5}\right)$, since $\sqrt{4^2+3^2}=5$.
💡 Perpendicular lines have slopes that multiply to $-1$, so flipping and negating $-\tfrac{4}{3}$ gives the direction $F$ must travel from $D$.
8.G.B.8 Step 4 Ride distance t along DE
- Let $t=DF$.
- Starting at $D=(0,0)$ and stepping distance $t$ along the unit direction $\left(\tfrac{4}{5},\tfrac{3}{5}\right)$ lands at $F=\left(\tfrac{4t}{5},\tfrac{3t}{5}\right)$.
- This really is distance $t$ from $D$ because $\sqrt{(4t/5)^2+(3t/5)^2}=\tfrac{t}{5}\sqrt{16+9}=t$.
💡 Because $(4,3)$ has length $5$, scaling it by $t/5$ moves you exactly $t$ units, so the variable $t$ is literally the distance we want.
8.EE.C.7 Step 5 Turn the right angle at F into an equation
- The condition $\overline{AF}\perp\overline{BF}$ means the legs $\vec{FA}$ and $\vec{FB}$ meet at a right angle, so their slopes multiply to $-1$ (equivalently $\vec{FA}\cdot\vec{FB}=0$).
- With $A=(0,12)$ and $B=(-5,0)$: $\left(\tfrac{4t}{5}-0\right)\left(\tfrac{4t}{5}+5\right)+\left(\tfrac{3t}{5}-12\right)\left(\tfrac{3t}{5}-0\right)=0$.
- Expanding: $\tfrac{16t^2}{25}+4t+\tfrac{9t^2}{25}-\tfrac{36t}{5}=0$, and $\tfrac{16t^2+9t^2}{25}=t^2$, so this collapses to $t^2-\tfrac{16t}{5}=0$.
💡 A right angle at $F$ says the two legs are perpendicular, which is exactly a zero dot product — one clean equation in the single unknown $t$.
6.NS.B.4 Step 6 Solve and pick the valid point
- Factor: $t\left(t-\tfrac{16}{5}\right)=0$, so $t=0$ or $t=\tfrac{16}{5}$.
- The root $t=0$ is the point $D$ itself (indeed $\angle ADB=90^\circ$ since $\overline{AD}\perp\overline{BC}$, so $D$ also sees $AB$ at a right angle) — but $F$ must be distinct from $D$.
- Also $DE=\tfrac{36}{5}=7.2>\tfrac{16}{5}=3.2$, so $F$ genuinely lies on segment $\overline{DE}$.
- Hence $DF=\tfrac{16}{5}$.
- Since $\gcd(16,5)=1$, we have $m=16$, $n=5$, and $m+n=21$, which is choice (B).
💡 The equation offers two right-angle points; we throw out $D$ (the forbidden duplicate) and keep the real $F$, then reduce the fraction to read off $m+n$.
8.G.B.7 The $13$-$14$-$15$ triangle splits neatly along its altitude to $BC$. If $AD=h$ 5.G.A.2 Place $D$ at the origin with $BC$ along the $x$-axis. Since $BD=5$ and $DC=9$, p 8.EE.B.6 Line $AC$ runs from $A=(0,12)$ to $C=(9,0)$, so its slope is $\frac{0-12}{9-0}=- 8.G.B.8 Let $t=DF$. Starting at $D=(0,0)$ and stepping distance $t$ along the unit direc 8.EE.C.7 The condition $\overline{AF}\perp\overline{BF}$ means the legs $\vec{FA}$ and $\ 6.NS.B.4 Factor: $t\left(t-\tfrac{16}{5}\right)=0$, so $t=0$ or $t=\tfrac{16}{5}$. The ro Review
Reasonableness: The two roots $t=0$ and $t=\tfrac{16}{5}$ have a clean meaning: both $D$ and $F$ lie on the circle with diameter $AB$ (every point on that circle sees $AB$ at $90^\circ$), and line $DE$ cuts that circle in exactly those two points. Rejecting $D$ leaves $F$ at distance $\tfrac{16}{5}=3.2$ from $D$, comfortably inside $DE=\tfrac{36}{5}=7.2$, so $F$ sits on the segment as required. The fraction $\tfrac{16}{5}$ is already reduced, giving $m+n=21$ — answer (B), which is on the list.
Alternative: Tool #7 (Identify Subproblems) with pure circle geometry: since $\angle ADB=\angle AFB=90^\circ$, points $A,F,D,B$ are concyclic on the circle with diameter $AB$. Reading $DF$ as a chord and using the similar right triangles ($\triangle ADF\sim\triangle$ pieces of the figure) or Ptolemy's Theorem on $ADBF$ recovers the same $DF=\tfrac{16}{5}$ without coordinates — but the coordinate route turns the whole chain of perpendiculars into one short equation.
CCSS standards used (min grade 8)
8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Finding $BD=5$, $DC=9$, and the altitude $AD=12$ in the $13$-$14$-$15$ triangle.)5.G.A.2Represent real-world and mathematical problems by graphing points (Placing $D$, $B$, $C$, and $A$ on coordinate axes with the altitude on the $y$-axis.)8.EE.B.6Use similar triangles to explain why the slope is the same between any two points (Getting slope $AC=-\tfrac{4}{3}$ and the perpendicular slope $\tfrac{3}{4}$ for line $DE$.)8.G.B.8Apply the Pythagorean theorem to find distance between two points in a coordinate system (Verifying that stepping $t/5$ along $(4,3)$ moves exactly distance $t$, so $DF=t$.)8.EE.C.7Solve linear equations in one variable (Setting the dot product to zero and solving $t^2-\tfrac{16t}{5}=0$ for $t$.)6.NS.B.4Find greatest common factor and least common multiple of two numbers (Confirming $\gcd(16,5)=1$ so $\tfrac{16}{5}$ is in lowest terms and $m+n=21$.)
⭐ Pin the $13$-$14$-$15$ triangle to axes so every 'perpendicular' becomes a slope; walk distance $t$ up line $DE$ to reach $F$, and the right angle at $F$ gives one equation $t^2-\tfrac{16}{5}t=0$. Toss out the root that is just point $D$, keep $DF=\tfrac{16}{5}$, so $m+n=21$ — choice (B). It looks like a hard geometry problem, but coordinates turn it into Grade 8 algebra.
⭐ Pin the $13$-$14$-$15$ triangle to axes so every 'perpendicular' becomes a slope; walk distance $t$ up line $DE$ to reach $F$, and the right angle at $F$ gives one equation $t^2-\tfrac{16}{5}t=0$. Toss out the root that is just point $D$, keep $DF=\tfrac{16}{5}$, so $m+n=21$ — choice (B). It looks like a hard geometry problem, but coordinates turn it into Grade 8 algebra.
More like this
Same archetype — closest grade level first.
