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Difference between revisions of "2004 AIME II Problems/Problem 13"

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However, <math>h_{BDE} = h_{ABC} + h_{CAF} + h_{EFD}</math>, with all three heights oriented in the same direction. Since <math>\triangle ABC \cong \triangle CFA</math>, it follows that <math>h_{ABC} = h_{CAF}</math>, and from the similarity ratio, <math>h_{EFD} = \frac{15}{7}h_{ABC}</math>. Hence <math>\frac{h_{ABC}}{h_{BDE}} = \frac{h_{ABC}}{2h_{ABC} + \frac {15}7h_{ABC}} = \frac{7}{29}</math>, and the ratio of the areas is <math>\frac{7}{15} \cdot \frac 7{29} = \frac{49}{435}</math>. The answer is <math>m+n = \boxed{484}</math>.
 
However, <math>h_{BDE} = h_{ABC} + h_{CAF} + h_{EFD}</math>, with all three heights oriented in the same direction. Since <math>\triangle ABC \cong \triangle CFA</math>, it follows that <math>h_{ABC} = h_{CAF}</math>, and from the similarity ratio, <math>h_{EFD} = \frac{15}{7}h_{ABC}</math>. Hence <math>\frac{h_{ABC}}{h_{BDE}} = \frac{h_{ABC}}{2h_{ABC} + \frac {15}7h_{ABC}} = \frac{7}{29}</math>, and the ratio of the areas is <math>\frac{7}{15} \cdot \frac 7{29} = \frac{49}{435}</math>. The answer is <math>m+n = \boxed{484}</math>.
 +
 +
== Additional Trigonometry-Free Alternative ==
 +
 +
Instead of using the Law of Cosines, we can draw a line perpendicular to line BC down from point A until it intersects BC at a point <math>P</math>. Since <math>\angle PBA = 60^{\circ}</math>, we can use the <math>30-60-90</math> triangle to deduce that <math>PB = \frac{3}{2}</math>, and <math>PA = \frac{3\sqrt{3}}{2}</math>. From here, we can use Pythagorean theorem to deduce that <math>AC = 7</math>. Then, we can follow with the rest of the solution above.
  
 
== See also ==
 
== See also ==

Latest revision as of 21:15, 7 April 2023

Problem

Let $ABCDE$ be a convex pentagon with $AB \parallel CE, BC \parallel AD, AC \parallel DE, \angle ABC=120^\circ, AB=3, BC=5,$ and $DE = 15.$ Given that the ratio between the area of triangle $ABC$ and the area of triangle $EBD$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$

Solution

Let the intersection of $\overline{AD}$ and $\overline{CE}$ be $F$. Since $AB \parallel CE, BC \parallel AD,$ it follows that $ABCF$ is a parallelogram, and so $\triangle ABC \cong \triangle CFA$. Also, as $AC \parallel DE$, it follows that $\triangle ABC \sim \triangle EFD$.

[asy] pointpen = black; pathpen = black+linewidth(0.7); pair D=(0,0), E=(15,0), F=IP(CR(D, 75/7), CR(E, 45/7)), A=D+ (5+(75/7))/(75/7) * (F-D), C = E+ (3+(45/7))/(45/7) * (F-E), B=IP(CR(A,3), CR(C,5)); D(MP("A",A,(1,0))--MP("B",B,N)--MP("C",C,NW)--MP("D",D)--MP("E",E)--cycle); D(D--A--C--E); D(MP("F",F)); MP("5",(B+C)/2,NW); MP("3",(A+B)/2,NE); MP("15",(D+E)/2); [/asy]

By the Law of Cosines, $AC^2 = 3^2 + 5^2 - 2 \cdot 3 \cdot 5 \cos 120^{\circ} = 49 \Longrightarrow AC = 7$. Thus the length similarity ratio between $\triangle ABC$ and $\triangle EFD$ is $\frac{AC}{ED} = \frac{7}{15}$.

Let $h_{ABC}$ and $h_{BDE}$ be the lengths of the altitudes in $\triangle ABC, \triangle BDE$ to $AC, DE$ respectively. Then, the ratio of the areas $\frac{[ABC]}{[BDE]} = \frac{\frac 12 \cdot h_{ABC} \cdot AC}{\frac 12 \cdot h_{BDE} \cdot DE} = \frac{7}{15} \cdot \frac{h_{ABC}}{h_{BDE}}$.

However, $h_{BDE} = h_{ABC} + h_{CAF} + h_{EFD}$, with all three heights oriented in the same direction. Since $\triangle ABC \cong \triangle CFA$, it follows that $h_{ABC} = h_{CAF}$, and from the similarity ratio, $h_{EFD} = \frac{15}{7}h_{ABC}$. Hence $\frac{h_{ABC}}{h_{BDE}} = \frac{h_{ABC}}{2h_{ABC} + \frac {15}7h_{ABC}} = \frac{7}{29}$, and the ratio of the areas is $\frac{7}{15} \cdot \frac 7{29} = \frac{49}{435}$. The answer is $m+n = \boxed{484}$.

Additional Trigonometry-Free Alternative

Instead of using the Law of Cosines, we can draw a line perpendicular to line BC down from point A until it intersects BC at a point $P$. Since $\angle PBA = 60^{\circ}$, we can use the $30-60-90$ triangle to deduce that $PB = \frac{3}{2}$, and $PA = \frac{3\sqrt{3}}{2}$. From here, we can use Pythagorean theorem to deduce that $AC = 7$. Then, we can follow with the rest of the solution above.

See also

2004 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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