Difference between revisions of "1971 AHSME Problems/Problem 28"

Latest revision as of 11:02, 7 August 2024

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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
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Difference between revisions of "1971 AHSME Problems/Problem 28"

Latest revision as of 11:02, 7 August 2024

Problem

Solution

See Also

@@ Line 11: / Line 11: @@
 == Solution ==
-<math>\boxed{\textbf{(C) }200}</math>.
+<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 <math>\triangle ABC</math> with base <math>\overline{BC}</math> and longest parallel segment <math>\overline{DE}</math> with <math>D</math> on <math>\overline{AB}</math> and <math>E</math> on <math>\overline{AC}</math>, as in the diagram.
+By the properties of [[transversals]], we have <math>\measuredangle ABC = \measuredangle ADE</math>. Thus, by [[AA Similarity]], we have <math>\triangle ADE \sim \triangle ABC</math> (because they share <math>\angle BAC</math>). From the problem, we know that <math>\tfrac{AE}{AC}=\tfrac9{10}</math>, so, by similarity, <math>\tfrac{DE}{BC}=\tfrac9{10}</math>, and so <math>DE=\tfrac9{10}BC</math>.
+Now, let <math>[\triangle ABC] = A</math>. Because <math>\tfrac{DE}{BE}=\tfrac9{10}</math>, we know that <math>\tfrac{[\triangle ADE]}{[\triangle ABC]}=(\tfrac9{10})^2=\tfrac{81}{100}</math>. From the problem, <math>[BCED]=38</math>, so <math>A=[\triangle ADE]+38=\tfrac9{10}A+38</math>. Solving for <math>A</math> yields  <math>A=\boxed{\textbf{(C) }200}</math>.
 == See Also ==
 {{AHSME 35p box|year=1971|num-b=27|num-a=29}}
 {{MAA Notice}}
+[[Category:Introductory Geometry Problems]]