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Difference between revisions of "1990 AHSME Problems/Problem 20"

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== Solution ==
 
== Solution ==
−
<math>\fbox{C}</math>
+
 
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Label the angles as shown in the diagram. Since <math>\angle DEC</math> forms a linear pair with <math>\angle DEA</math>, <math>\angle DEA</math> is a right angle.
 +
 
 +
<asy>
 +
pair A = (0,0), B = (7,4.2), C = (10, 0), D = (3, -5), E = (3, 0), F = (7,0);
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draw(A--B--C--D--cycle,dot);
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draw(A--E--F--C,dot);
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draw(D--E--F--B,dot);
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 +
markscalefactor = 0.075;
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draw(rightanglemark(B, A, D));
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draw(rightanglemark(D, E, A));
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draw(rightanglemark(B, F, A));
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draw(rightanglemark(D, C, B));
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draw(rightanglemark(D, E, C));
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draw(rightanglemark(B, F, C));
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MP("A",(0,0),W);
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MP("B",(7,4.2),N);
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MP("C",(10,0),E);
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MP("D",(3,-5),S);
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MP("E",(3,0),N);
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MP("F",(7,0),S);
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</asy>
 +
 
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Let <math>\angle DAE = \angle \alpha</math> and <math>\angle BAF = \angle \beta</math>.
 +
 
 +
Since <math>\angle \alpha + \angle \beta = 90^\circ</math>, and <math>\angle \alpha + \angle BAF = 90^\circ</math>, then <math>\angle \beta = \angle BAF</math>. By the same logic, <math>\angle ABF = \angle \alpha</math>.
 +
 
 +
 
 +
As a result, <math>\triangle AED \sim \triangle BFA</math>. By the same logic, <math>\triangle CFB \sim \triangle DEC</math>.
 +
 
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Then, <math>\frac{BF}{AF} = \frac{3}{5}</math>, and <math>\frac{CF}{BF} = \frac{5}{7}</math>.
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Then, <math>7CF = 5BF</math>, and <math>5BF = 3AF</math>.
 +
 
 +
By the transitive property, <math>7CF = 3AF</math>. <math>AC = AF + CF = 10</math>, and plugging in, we get <math>CF = 3</math>.
 +
 
 +
Finally, plugging in to <math>\frac{CF}{BF} = \frac{5}{7}</math>, we get <math>BF = 4.2</math> <math>\fbox{C}</math>
  
 
== See also ==
 
== See also ==

Revision as of 12:55, 22 February 2018

Problem

[asy] pair A = (0,0), B = (7,4.2), C = (10, 0), D = (3, -5), E = (3, 0), F = (7,0); draw(A--B--C--D--cycle,dot); draw(A--E--F--C,dot); draw(D--E--F--B,dot); markscalefactor = 0.1; draw(rightanglemark(B, A, D)); draw(rightanglemark(D, E, C)); draw(rightanglemark(B, F, A)); draw(rightanglemark(D, C, B)); MP("A",(0,0),W); MP("B",(7,4.2),N); MP("C",(10,0),E); MP("D",(3,-5),S); MP("E",(3,0),N); MP("F",(7,0),S); [/asy]

In the figure $ABCD$ is a quadrilateral with right angles at $A$ and $C$. Points $E$ and $F$ are on $\overline{AC}$, and $\overline{DE}$ and $\overline{BF}$ are perpendicual to $\overline{AC}$. If $AE=3, DE=5,$ and $CE=7$, then $BF=$

$\text{(A) } 3.6\quad \text{(B) } 4\quad \text{(C) } 4.2\quad \text{(D) } 4.5\quad \text{(E) } 5$

Solution

Label the angles as shown in the diagram. Since $\angle DEC$ forms a linear pair with $\angle DEA$, $\angle DEA$ is a right angle.

[asy] pair A = (0,0), B = (7,4.2), C = (10, 0), D = (3, -5), E = (3, 0), F = (7,0); draw(A--B--C--D--cycle,dot); draw(A--E--F--C,dot); draw(D--E--F--B,dot); markscalefactor = 0.075; draw(rightanglemark(B, A, D)); draw(rightanglemark(D, E, A)); draw(rightanglemark(B, F, A)); draw(rightanglemark(D, C, B)); draw(rightanglemark(D, E, C)); draw(rightanglemark(B, F, C)); MP("A",(0,0),W); MP("B",(7,4.2),N); MP("C",(10,0),E); MP("D",(3,-5),S); MP("E",(3,0),N); MP("F",(7,0),S); [/asy]

Let $\angle DAE = \angle \alpha$ and $\angle BAF = \angle \beta$.

Since $\angle \alpha + \angle \beta = 90^\circ$, and $\angle \alpha + \angle BAF = 90^\circ$, then $\angle \beta = \angle BAF$. By the same logic, $\angle ABF = \angle \alpha$.


As a result, $\triangle AED \sim \triangle BFA$. By the same logic, $\triangle CFB \sim \triangle DEC$.

Then, $\frac{BF}{AF} = \frac{3}{5}$, and $\frac{CF}{BF} = \frac{5}{7}$.

Then, $7CF = 5BF$, and $5BF = 3AF$.

By the transitive property, $7CF = 3AF$. $AC = AF + CF = 10$, and plugging in, we get $CF = 3$.

Finally, plugging in to $\frac{CF}{BF} = \frac{5}{7}$, we get $BF = 4.2$ $\fbox{C}$

See also

1990 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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
All AHSME Problems and Solutions

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