AMC 10 · 2007 · #17

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Polygon Triangulation

equilateral-trianglearea-triangles double-counting ↑ Prerequisites: area-triangles

📏 Medium solution 💡 3 insights

Problem

Point

P

sits inside an equilateral triangle

ABC

. Dropping perpendiculars from

P

to the three sides gives feet

Q

AB

R

BC

, and

S

CA

, with lengths

PQ=1

PR=2

, and

PS=3

. Find the side length

AB

Pick an answer.

(A)

$\ 4$

(B)

$\ 3\sqrt{3}$

(C)

$\ 6$

(D)

$\ 4\sqrt{3}$

(E)

$\ 9$

AMC 10 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

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.

1STEP 1

Split the triangle from P

Draw $PA$ , $PB$ , $PC$ : the triangle splits into three pieces, each with a full side as base and one given distance as height.

[ABC]=[PAB]+[PBC]+[PCA]

2STEP 2

Add the three small areas

Every piece has base $s$ , so the areas $\frac{1}{2}s(1)$ , $\frac{1}{2}s(2)$ , $\frac{1}{2}s(3)$ add to $3s$ .

[ABC]=1/2 s(1)+1/2 s(2)+1/2 s(3)=1/2 s(1+2+3)=3s

3STEP 3

Area of the whole equilateral triangle

Cut the equilateral triangle down its own altitude $\frac{\sqrt{3}}{2}s$ (two 30-60-90 halves), giving area $\frac{\sqrt{3}}{4}s^2$ .

h=√(s²-s²/4)=√3/2s, [ABC]=√3/4s²

4STEP 4

Set the two areas equal

One region, two measurements: $3s=\frac{\sqrt{3}}{4}s^2$ . Dividing by the positive $s$ leaves $s=\frac{12}{\sqrt{3}}$ .

3s=√3/4s² → 3=√3/4s → s=12/√3

5STEP 5

Simplify the radical

Multiply top and bottom by $\sqrt{3}$ to get $\frac{12\sqrt{3}}{3}$ , so $AB=4\sqrt{3}$ — choice (D).

s=12/√3=12√3/3=4√3

Answer

4√(3)

Check both areas with s=4√3. The pieces give 3s=12√3. The whole gives √3/4s²=√3/4 · 48=12√3 — they agree. Also note the altitude of the whole triangle is h=√3/2 · 4√3=6, exactly the sum 1+2+3 of the three distances; this is Viviani's Theorem, a reassuring cross-check. Numerically 4√3≈6.9, comfortably larger than the biggest perpendicular (3) as any real triangle must be, and it is the only choice matching 12/√3.

💡Key takeaway

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.

Split the triangle from P
Add the three small areas
Area of the whole equilateral triangle
Set the two areas equal
Simplify the radical

Split the triangle from P

Add the three small areas

Area of the whole equilateral triangle

Set the two areas equal

Simplify the radical

More like this