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Difference between revisions of "1984 AIME Problems/Problem 3"

m (+img)
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== Problem ==
 
== Problem ==
−
A [[point]] <math>\displaystyle P</math> is chosen in the interior of <math>\displaystyle \triangle ABC</math> such that when [[line]]s are drawn through <math>\displaystyle P</math> [[parallel]] to the sides of <math>\displaystyle \triangle ABC</math>, the resulting smaller [[triangle]]s <math>\displaystyle t_{1}</math>, <math>\displaystyle t_{2}</math>, and <math>\displaystyle t_{3}</math> in the figure, have [[area]]s <math>\displaystyle 4</math>, <math>\displaystyle 9</math>, and <math>\displaystyle 49</math>, respectively. Find the area of <math>\displaystyle \triangle ABC</math>.
+
A [[point]] <math>P</math> is chosen in the interior of <math>\triangle ABC</math> such that when [[line]]s are drawn through <math>P</math> [[parallel]] to the sides of <math>\triangle ABC</math>, the resulting smaller [[triangle]]s <math>t_{1}</math>, <math>t_{2}</math>, and <math>t_{3}</math> in the figure, have [[area]]s <math>4</math>, <math>9</math>, and <math>49</math>, respectively. Find the area of <math>\triangle ABC</math>.
−
[[Image:1984_AIME-3.png]]
+
<center><asy>
 +
size(200);
 +
pathpen=black;pointpen=black;
 +
pair A=(0,0),B=(12,0),C=(4,5);
 +
D(A--B--C--cycle); D(A+(B-A)*3/4--A+(C-A)*3/4); D(B+(C-B)*5/6--B+(A-B)*5/6);D(C+(B-C)*5/12--C+(A-C)*5/12);
 +
MP("A",C,N);MP("B",A,SW);MP("C",B,SE); /* sorry mixed up points according to resources diagram. */
 +
MP("t_3",(A+B+(B-A)*3/4+(A-B)*5/6)/2+(-1,0.8),N);
 +
MP("t_2",(B+C+(B-C)*5/12+(C-B)*5/6)/2+(-0.3,0.1),WSW);
 +
MP("t_1",(A+C+(C-A)*3/4+(A-C)*5/12)/2+(0,0.15),ESE);
 +
</asy></center>
  
 
== Solution ==
 
== Solution ==
−
Notice that all four triangles are [[similar triangles|similar]] to each other. Also, note that the length of any one side of the larger triangle is equation to the sum of the sides of each of the corresponding sides on the smaller triangles. Since the squares of the lengths of the sides are in proportion with the area, we can write the lengths of corresponding sides of the triangle as <math>2x,\ 3x,\ 7x</math>. Thus, the corresponding side on the large triangle is <math>12x</math>, and the area of the triangle is <math>12^2 = 144</math>.
+
Notice that all four triangles are [[similar triangles|similar]] to each other. Also, note that the length of any one side of the larger triangle is equation to the sum of the sides of each of the corresponding sides on the smaller triangles. Since the squares of the lengths of the sides are in proportion with the area, we can write the lengths of corresponding sides of the triangle as <math>2x,\ 3x,\ 7x</math>. Thus, the corresponding side on the large triangle is <math>12x</math>, and the area of the triangle is <math>12^2 = \boxed{144}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1984|num-b=2|num-a=4}}
 
{{AIME box|year=1984|num-b=2|num-a=4}}
−
* [[AIME Problems and Solutions]]
 
−
* [[American Invitational Mathematics Examination]]
 
−
* [[Mathematics competition resources]]
 
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Revision as of 17:18, 9 April 2008

Problem

A point $P$ is chosen in the interior of $\triangle ABC$ such that when lines are drawn through $P$ parallel to the sides of $\triangle ABC$, the resulting smaller triangles $t_{1}$, $t_{2}$, and $t_{3}$ in the figure, have areas $4$, $9$, and $49$, respectively. Find the area of $\triangle ABC$.

[asy] size(200); pathpen=black;pointpen=black; pair A=(0,0),B=(12,0),C=(4,5); D(A--B--C--cycle); D(A+(B-A)*3/4--A+(C-A)*3/4); D(B+(C-B)*5/6--B+(A-B)*5/6);D(C+(B-C)*5/12--C+(A-C)*5/12); MP("A",C,N);MP("B",A,SW);MP("C",B,SE); /* sorry mixed up points according to resources diagram. */ MP("t_3",(A+B+(B-A)*3/4+(A-B)*5/6)/2+(-1,0.8),N); MP("t_2",(B+C+(B-C)*5/12+(C-B)*5/6)/2+(-0.3,0.1),WSW); MP("t_1",(A+C+(C-A)*3/4+(A-C)*5/12)/2+(0,0.15),ESE); [/asy]

Solution

Notice that all four triangles are similar to each other. Also, note that the length of any one side of the larger triangle is equation to the sum of the sides of each of the corresponding sides on the smaller triangles. Since the squares of the lengths of the sides are in proportion with the area, we can write the lengths of corresponding sides of the triangle as $2x,\ 3x,\ 7x$. Thus, the corresponding side on the large triangle is $12x$, and the area of the triangle is $12^2 = \boxed{144}$.

See also

1984 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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