1971 AHSME Problems/Problem 28

Problem

Nine lines parallel to the base of a triangle divide the other sides each into $10$ equal segments and the area into $10$ distinct parts. If the area of the largest of these parts is $38$ , then the area of the original triangle is

$\textbf{(A) }180\qquad \textbf{(B) }190\qquad \textbf{(C) }200\qquad \textbf{(D) }210\qquad \textbf{(E) }240$

Solution

$[asy] import geometry; point B = origin; point A = (3,5); point C = (7,0); triangle t = triangle(A,B,C); point D = B*9/10 + A/10; point E; // Defining point E pair[] e = intersectionpoints(parallel(D,line(B,C)),A--C); E = e[0]; // Triangle ABC and Parallel Segment draw(t); draw(D--E); // Point Labels dot(A); label("A",A,NW); dot(B); label("B",B,SW); dot(C); label("C",C,SE); dot(D); label("D",D,NW); dot(E); label("E",E,NE); [/asy]$

Let the triangle be $\triangle ABC$ with base $\overline{BC}$ and longest parallel segment $\overline{DE}$ with $D$ on $\overline{AB}$ and $E$ on $\overline{AC}$ , as in the diagram.

By the properties of transversals, we have $\measuredangle ABC = \measuredangle ADE$ . Thus, by AA Similarity, we have $\triangle ADE \sim \triangle ABC$ (because they share $\angle BAC$ ). From the problem, we know that $\tfrac{AE}{AC}=\tfrac9{10}$ , so, by similarity, $\tfrac{DE}{BC}=\tfrac9{10}$ , and so $DE=\tfrac9{10}BC$ .

Now, let $[\triangle ABC] = A$ . Because $\tfrac{DE}{BE}=\tfrac9{10}$ , we know that $\tfrac{[\triangle ADE]}{[\triangle ABC]}=(\tfrac9{10})^2=\tfrac{81}{100}$ . From the problem, $[BCED]=38$ , so $A=[\triangle ADE]+38=\tfrac9{10}A+38$ . Solving for $A$ yields $A=\boxed{\textbf{(C) }200}$ .

1971 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 27	Followed by Problem 29
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1971 AHSME Problems/Problem 28

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