AMC 10 · 2007 · #17
Grade 8 geometry-2dPoint P is inside equilateral △ABC. Points Q, R, and S are the feet of the perpendiculars from P to AB, BC, and CA, respectively. Given that PQ=1, PR=2, and PS=3, what is AB?
Pick an answer.
AMC 10 2007 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: Point $P$ sits inside an equilateral triangle $ABC$. Dropping perpendiculars from $P$ to the three sides gives feet $Q$ on $\overline{AB}$, $R$ on $\overline{BC}$, and $S$ on $\overline{CA}$, with lengths $PQ=1$, $PR=2$, and $PS=3$. Find the side length $AB$.
Givens: $\triangle ABC$ is equilateral, so all three sides have the same length and every angle is $60^\circ$; $P$ is an interior point, and $PQ,PR,PS$ are its perpendicular distances to the three sides: $PQ=1$, $PR=2$, $PS=3$; Answer choices: (A) $4$, (B) $3\sqrt{3}$, (C) $6$, (D) $4\sqrt{3}$, (E) $9$
Unknowns: The side length $AB$ of the equilateral triangle
Understand
Restated: Point $P$ sits inside an equilateral triangle $ABC$. Dropping perpendiculars from $P$ to the three sides gives feet $Q$ on $\overline{AB}$, $R$ on $\overline{BC}$, and $S$ on $\overline{CA}$, with lengths $PQ=1$, $PR=2$, and $PS=3$. Find the side length $AB$.
Givens: $\triangle ABC$ is equilateral, so all three sides have the same length and every angle is $60^\circ$; $P$ is an interior point, and $PQ,PR,PS$ are its perpendicular distances to the three sides: $PQ=1$, $PR=2$, $PS=3$; Answer choices: (A) $4$, (B) $3\sqrt{3}$, (C) $6$, (D) $4\sqrt{3}$, (E) $9$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #4 Introduce a Variable, #13 Convert to Algebra
The three perpendiculars beg to be joined into a picture: connect $P$ to each vertex and the equilateral triangle splits into three smaller triangles (Tool #1, then Tool #7). Because each perpendicular is the height of one small triangle onto a side of length $s=AB$, the three areas are easy to write down and add. Setting that total equal to the area of the whole equilateral triangle — computed with one variable $s$ (Tool #4) — turns the geometry into a single equation to solve (Tool #13). This area-splitting idea is exactly Viviani's Theorem in disguise.
Execute — Answer: D
6.G.A.1 Step 1 Split the triangle from $P$
- Draw segments $PA$, $PB$, and $PC$.
- These cut $\triangle ABC$ into three smaller triangles: $\triangle PAB$, $\triangle PBC$, and $\triangle PCA$.
- Each perpendicular is the height of one of these triangles dropped onto a full side: $PQ\perp AB$ is the height of $\triangle PAB$, $PR\perp BC$ the height of $\triangle PBC$, and $PS\perp CA$ the height of $\triangle PCA$.
- Every base is a side of the equilateral triangle, so each base has the same length $s=AB$.
💡 Joining the inside point to the corners turns one triangle into three whose heights are exactly the numbers you were handed.
7.G.B.6 Step 2 Add the three small areas
- Area of a triangle is $\tfrac12\,(\text{base})(\text{height})$.
- Each small triangle has base $s$ and one of the given heights: $[PAB]=\tfrac12 s\cdot 1$, $[PBC]=\tfrac12 s\cdot 2$, $[PCA]=\tfrac12 s\cdot 3$.
- Adding them factors out $\tfrac12 s$: the total is $\tfrac12 s(1+2+3)=3s$.
💡 Sharing the same base $s$ lets the three heights simply add, so only their sum $1+2+3$ matters.
8.G.B.7 Step 3 Area of the whole equilateral triangle
- Now measure $[ABC]$ directly.
- Split the equilateral triangle down its own altitude into two $30$-$60$-$90$ right triangles with hypotenuse $s$ and half-base $\tfrac{s}{2}$.
- By the Pythagorean theorem the altitude is $h=\sqrt{s^2-\left(\tfrac{s}{2}\right)^2}=\tfrac{\sqrt3}{2}\,s$.
- So $[ABC]=\tfrac12\, s\cdot h=\tfrac12 s\cdot\tfrac{\sqrt3}{2}s=\tfrac{\sqrt3}{4}s^2$.
💡 An equilateral triangle folds into two $30$-$60$-$90$ triangles, so its area is always $\tfrac{\sqrt3}{4}$ times the side squared.
7.EE.B.4 Step 4 Set the two areas equal
- Both expressions describe the same region, so they must match: $3s=\tfrac{\sqrt3}{4}s^2$.
- Since $s>0$, divide both sides by $s$ to get $3=\tfrac{\sqrt3}{4}s$.
- Multiplying by $\tfrac{4}{\sqrt3}$ isolates $s$: $s=\tfrac{12}{\sqrt3}$.
💡 One region measured two ways gives an equation, and dividing out the shared $s$ collapses it to a quick linear solve.
8.EE.A.2 Step 5 Simplify the radical
- Rationalize $\tfrac{12}{\sqrt3}$ by multiplying top and bottom by $\sqrt3$: $\tfrac{12}{\sqrt3}\cdot\tfrac{\sqrt3}{\sqrt3}=\tfrac{12\sqrt3}{3}=4\sqrt3$.
- So $AB=4\sqrt3$, which is choice $\textbf{(D)}$.
💡 Clearing the root from the denominator turns the messy fraction into the clean form $4\sqrt3$.
6.G.A.1 Draw segments $PA$, $PB$, and $PC$. These cut $\triangle ABC$ into three smaller 7.G.B.6 Area of a triangle is $\tfrac12\,(\text{base})(\text{height})$. Each small trian 8.G.B.7 Now measure $[ABC]$ directly. Split the equilateral triangle down its own altitu 7.EE.B.4 Both expressions describe the same region, so they must match: $3s=\tfrac{\sqrt3 8.EE.A.2 Rationalize $\tfrac{12}{\sqrt3}$ by multiplying top and bottom by $\sqrt3$: $\tf Review
Reasonableness: Check both areas with $s=4\sqrt3$. The pieces give $3s=12\sqrt3$. The whole gives $\tfrac{\sqrt3}{4}s^2=\tfrac{\sqrt3}{4}\cdot 48=12\sqrt3$ — they agree. Also note the altitude of the whole triangle is $h=\tfrac{\sqrt3}{2}\cdot 4\sqrt3=6$, exactly the sum $1+2+3$ of the three distances; this is Viviani's Theorem, a reassuring cross-check. Numerically $4\sqrt3\approx6.9$, comfortably larger than the biggest perpendicular ($3$) as any real triangle must be, and it is the only choice matching $12/\sqrt3$.
Alternative: Skip the area bookkeeping and use Viviani's Theorem directly: for any interior point of an equilateral triangle, the perpendicular distances to the three sides always add up to the triangle's altitude. So the altitude is $h=1+2+3=6$. Since $h=\tfrac{\sqrt3}{2}s$ for an equilateral triangle, solve $6=\tfrac{\sqrt3}{2}s$ to get $s=\tfrac{12}{\sqrt3}=4\sqrt3$.
CCSS standards used (min grade 8)
6.G.A.1Find the area of triangles and other figures by composing and decomposing (Splitting $\triangle ABC$ into $\triangle PAB$, $\triangle PBC$, $\triangle PCA$ and seeing each perpendicular as a height.)7.G.B.6Solve problems involving area of two-dimensional objects composed of triangles (Adding the three sub-triangle areas $\tfrac12 s(1)+\tfrac12 s(2)+\tfrac12 s(3)=3s$.)8.G.B.7Apply the Pythagorean Theorem to determine unknown side lengths (Finding the equilateral triangle's altitude $h=\tfrac{\sqrt3}{2}s$ to get area $\tfrac{\sqrt3}{4}s^2$.)7.EE.B.4Use variables to construct and solve simple equations (Setting $3s=\tfrac{\sqrt3}{4}s^2$ and solving for $s$.)8.EE.A.2Use square root symbols and work with irrational square roots (Rationalizing $\tfrac{12}{\sqrt3}$ into the simplified form $4\sqrt3$.)
⭐ Connect the inside point to the corners: the three perpendiculars become heights, and matching the added-up pieces to the whole triangle's area pins down the side.
⭐ Connect the inside point to the corners: the three perpendiculars become heights, and matching the added-up pieces to the whole triangle's area pins down the side.
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