Difference between revisions of "1984 AIME Problems/Problem 3"
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== Problem == | == Problem == | ||
− | A point <math>P</math> is chosen in the interior of <math>\triangle ABC</math> such that when | + | 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>. |
+ | <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 1 == | ||
+ | By the transversals that go through <math>P</math>, all four triangles are [[similar triangles|similar]] to each other by the <math>AA</math> postulate. Also, note that the length of any one side of the larger triangle is equal to the sum of the sides of each of the corresponding sides on the smaller triangles. We use the identity <math>K = \dfrac{ab\sin C}{2}</math> to show that the areas are proportional (the sides are proportional and the angles are equal) Hence, 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>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | Alternatively, since the triangles are similar by <math>AA</math>, then the ratios between the bases and the heights of each of the three triangles would all be equal. The areas of each of the triangles are all perfect squares, so we could assume <math>\dfrac{base}{height} = \dfrac{2}{1}.</math> That means that the base of <math>t_{1}</math> is 4, the base of <math>t_{2}</math> is 6, and the base of <math>t_{3}</math> is 14. Since the quadrilaterals underneath <math>t_{1}</math> and <math>t_{2}</math> are both parallelograms, and opposite sides of a parallelogram are congruent, the base of the large triangle is <math>4 + 14 + 6 = 24</math>. Therefore, the height of the entire triangle would be twelve, so therefore, the area of the large triangle is <math>\dfrac{1}{2} \cdot 24 \cdot 12 = \boxed{144}</math>. | ||
+ | |||
+ | == Solution 3 == | ||
+ | The base of <math>\triangle{ABC}</math> is <math>BC</math>. Let the base of <math>t_1</math> be <math>x</math>, the base of <math>t_2</math> be <math>y</math>, and the base of <math>t_3</math> be <math>z</math>. Since <math>\triangle{ABC}, t_1, t_2,</math> and <math>t_3</math> are all similar, the sections in <math>\triangle{ABC}</math> that aren't <math>t_1,t_2,</math> or <math>t_3</math> are all parallelograms. Hence, <math>BC=x+z+y</math>. We can relate <math>t_1,t_2,</math> and <math>t_3</math> by the square root of the ratio of their areas. <math>\sqrt{\frac{4}{9}}=\frac{2}{3}</math> and <math>\sqrt{\frac{4}{49}}=\frac{2}{7}</math> so <math>y=\frac{3x}{2}</math> and <math>z=\frac{7x}{2}</math>. <math>x+\frac{7x}{2}+\frac{3x}{2}=6x</math>, so <math>\triangle{ABC}</math> has a base that is <math>6</math> times <math>t_1</math>. <math>[\triangle{ABC}]=36[t_1]=36 \cdot 4=\boxed{144}</math>. | ||
+ | |||
+ | -PhunsukhWangdu | ||
+ | |||
+ | |||
+ | |||
+ | == Solution 4 == | ||
+ | Since the three lines through <math>P</math> are parallel to the sides, <math>t_1</math>, <math>t_2</math>, <math>t_3</math>, and <math>\triangle{ABC}</math> are similar by <math>AA</math> similarity. Suppose the area of <math>\triangle{ABC}</math> is <math>x^2</math>, so the ratio of the base of <math>t_1</math> to the base of <math>t_2</math> to the base of <math>t_3</math> to the base of <math>\triangle{ABC}</math> is <math>2:3:7:x</math>. Because the quadrilaterals below <math>t_1</math> and <math>t_2</math> are parallelograms, the base of <math>\triangle{ABC}</math> is equal to the sum of the bases of <math>t_1, t_2,</math> and <math>t_3</math>. Therefore, <math>x</math> equals <math>2+3+7=12</math> so the area of <math>\triangle{ABC}</math> equals <math>x^2=12^2=\boxed{144}.</math> | ||
+ | |||
+ | -Yiyj1 | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/hDsoyvFWYxc?t=199 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
− | |||
− | |||
== See also == | == See also == | ||
− | + | {{AIME box|year=1984|num-b=2|num-a=4}} | |
− | + | ||
− | + | [[Category:Intermediate Geometry Problems]] |
Latest revision as of 12:39, 13 August 2023
Contents
[hide]Problem
A point is chosen in the interior of
such that when lines are drawn through
parallel to the sides of
, the resulting smaller triangles
,
, and
in the figure, have areas
,
, and
, respectively. Find the area of
.
![[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]](http://latex.artofproblemsolving.com/2/9/3/293009bb673bfe7d5b163c98428082f101000e4d.png)
Solution 1
By the transversals that go through , all four triangles are similar to each other by the
postulate. Also, note that the length of any one side of the larger triangle is equal to the sum of the sides of each of the corresponding sides on the smaller triangles. We use the identity
to show that the areas are proportional (the sides are proportional and the angles are equal) Hence, we can write the lengths of corresponding sides of the triangle as
. Thus, the corresponding side on the large triangle is
, and the area of the triangle is
.
Solution 2
Alternatively, since the triangles are similar by , then the ratios between the bases and the heights of each of the three triangles would all be equal. The areas of each of the triangles are all perfect squares, so we could assume
That means that the base of
is 4, the base of
is 6, and the base of
is 14. Since the quadrilaterals underneath
and
are both parallelograms, and opposite sides of a parallelogram are congruent, the base of the large triangle is
. Therefore, the height of the entire triangle would be twelve, so therefore, the area of the large triangle is
.
Solution 3
The base of is
. Let the base of
be
, the base of
be
, and the base of
be
. Since
and
are all similar, the sections in
that aren't
or
are all parallelograms. Hence,
. We can relate
and
by the square root of the ratio of their areas.
and
so
and
.
, so
has a base that is
times
.
.
-PhunsukhWangdu
Solution 4
Since the three lines through are parallel to the sides,
,
,
, and
are similar by
similarity. Suppose the area of
is
, so the ratio of the base of
to the base of
to the base of
to the base of
is
. Because the quadrilaterals below
and
are parallelograms, the base of
is equal to the sum of the bases of
and
. Therefore,
equals
so the area of
equals
-Yiyj1
Video Solution by OmegaLearn
https://youtu.be/hDsoyvFWYxc?t=199
~ pi_is_3.14
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 |