Difference between revisions of "2004 AIME II Problems/Problem 13"
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== Problem == | == Problem == | ||
− | Let <math> ABCDE </math> be a convex pentagon with <math> AB | + | Let <math> ABCDE </math> be a [[convex]] [[pentagon]] with <math> AB \parallel CE, BC \parallel AD, AC \parallel DE, \angle ABC=120^\circ, AB=3, BC=5, </math> and <math>DE = 15. </math> Given that the [[ratio]] between the area of triangle <math> ABC </math> and the area of triangle <math> EBD </math> is <math> m/n, </math> where <math> m </math> and <math> n </math> are relatively prime positive integers, find <math> m+n. </math> |
== Solution == | == Solution == | ||
− | {{ | + | Let the intersection of <math>\overline{AD}</math> and <math>\overline{CE}</math> be <math>F</math>. Since <math>AB \parallel CE, BC \parallel AD, </math> it follows that <math>ABCF</math> is a [[parallelogram]], and so <math>\triangle ABC \cong \triangle CFA</math>. Also, as <math>AC \parallel DE</math>, it follows that <math>\triangle ABC \sim \triangle EFD</math>. |
+ | |||
+ | <center><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></center> | ||
+ | |||
+ | By the [[Law of Cosines]], <math>AC^2 = 3^2 + 5^2 - 2 \cdot 3 \cdot 5 \cos 120^{\circ} = 49 \Longrightarrow AC = 7</math>. Thus the length similarity ratio between <math>\triangle ABC</math> and <math>\triangle EFD</math> is <math>\frac{AC}{ED} = \frac{7}{15}</math>. | ||
+ | |||
+ | Let <math>h_{ABC}</math> and <math>h_{BDE}</math> be the lengths of the [[altitude]]s in <math>\triangle ABC, \triangle BDE</math> to <math>AC, DE</math> respectively. Then, the ratio of the areas <math>\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}}</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 == | ||
{{AIME box|year=2004|n=II|num-b=12|num-a=14}} | {{AIME box|year=2004|n=II|num-b=12|num-a=14}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 22:15, 7 April 2023
Problem
Let be a convex pentagon with
and
Given that the ratio between the area of triangle
and the area of triangle
is
where
and
are relatively prime positive integers, find
Solution
Let the intersection of and
be
. Since
it follows that
is a parallelogram, and so
. Also, as
, it follows that
.
![[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]](http://latex.artofproblemsolving.com/9/c/d/9cd3a7b01b1890253f69e698559c0a908b73dde5.png)
By the Law of Cosines, . Thus the length similarity ratio between
and
is
.
Let and
be the lengths of the altitudes in
to
respectively. Then, the ratio of the areas
.
However, , with all three heights oriented in the same direction. Since
, it follows that
, and from the similarity ratio,
. Hence
, and the ratio of the areas is
. The answer is
.
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 . Since
, we can use the
triangle to deduce that
, and
. From here, we can use Pythagorean theorem to deduce that
. 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.