Difference between revisions of "2017 AMC 8 Problems/Problem 16"
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In the figure below, choose point <math>D</math> on <math>\overline{BC}</math> so that <math>\triangle ACD</math> and <math>\triangle ABD</math> have equal perimeters. What is the area of <math>\triangle ABD</math>? | In the figure below, choose point <math>D</math> on <math>\overline{BC}</math> so that <math>\triangle ACD</math> and <math>\triangle ABD</math> have equal perimeters. What is the area of <math>\triangle ABD</math>? | ||
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<asy>draw((0,0)--(4,0)--(0,3)--(0,0)); | <asy>draw((0,0)--(4,0)--(0,3)--(0,0)); | ||
label("$A$", (0,0), SW); | label("$A$", (0,0), SW); | ||
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<math>\textbf{(A) }\frac{3}{4}\qquad\textbf{(B) }\frac{3}{2}\qquad\textbf{(C) }2\qquad\textbf{(D) }\frac{12}{5}\qquad\textbf{(E) }\frac{5}{2}</math> | <math>\textbf{(A) }\frac{3}{4}\qquad\textbf{(B) }\frac{3}{2}\qquad\textbf{(C) }2\qquad\textbf{(D) }\frac{12}{5}\qquad\textbf{(E) }\frac{5}{2}</math> | ||
− | ==Solution 1== | + | ==Solutions== |
+ | |||
+ | ===Solution 1=== | ||
Essentially, we see that if we draw a line from point A to imaginary point D, that line would apply to both triangles. Let us say that <math>x</math> is the lenmeter of <math>\triangle{ABD}</math> would be <math>\overline{AD} + 4 + x</math>, while the perimeter of <math>\triangle{ACD}</math> would be <math>\overline{AD} + 3 + (5 - x)</math>. Notice that we can find out <math>x</math> from these two equations. We can find out that <math>x = 2</math>, and because the height of the triangles is the same, the ratio of the areas is <math>2:3</math>, so that means that the area of <math>\triangle ABD = \frac{2 \cdot 6}{5} = \boxed{\textbf{(D) } \frac{12}{5}}</math>. | Essentially, we see that if we draw a line from point A to imaginary point D, that line would apply to both triangles. Let us say that <math>x</math> is the lenmeter of <math>\triangle{ABD}</math> would be <math>\overline{AD} + 4 + x</math>, while the perimeter of <math>\triangle{ACD}</math> would be <math>\overline{AD} + 3 + (5 - x)</math>. Notice that we can find out <math>x</math> from these two equations. We can find out that <math>x = 2</math>, and because the height of the triangles is the same, the ratio of the areas is <math>2:3</math>, so that means that the area of <math>\triangle ABD = \frac{2 \cdot 6}{5} = \boxed{\textbf{(D) } \frac{12}{5}}</math>. | ||
− | ==Solution 2== | + | ===Solution 2=== |
We know that the perimeters of the two small triangles are <math>3+CD+AD</math> and <math>4+BD+AD</math>. Setting both equal and using <math>BD+CD = 5</math>, we have <math>BD = 2</math> and <math>CD = 3</math>. Now, we simply have to find the area of <math>\triangle ABD</math>. Since <math>\frac{BD}{CD} = \frac{2}{3}</math>, we must have <math>\frac{[ABD]}{[ACD]} = 2/3</math>. Combining this with the fact that <math>[ABC] = [ABD] + [ACD] = \frac{3\cdot4}{2} = 6</math>, we get <math>[ABD] = \frac{2}{5}[ABC] = \frac{2}{5} \cdot 6 = \boxed{\textbf{(D) } \frac{12}{5}}</math> | We know that the perimeters of the two small triangles are <math>3+CD+AD</math> and <math>4+BD+AD</math>. Setting both equal and using <math>BD+CD = 5</math>, we have <math>BD = 2</math> and <math>CD = 3</math>. Now, we simply have to find the area of <math>\triangle ABD</math>. Since <math>\frac{BD}{CD} = \frac{2}{3}</math>, we must have <math>\frac{[ABD]}{[ACD]} = 2/3</math>. Combining this with the fact that <math>[ABC] = [ABD] + [ACD] = \frac{3\cdot4}{2} = 6</math>, we get <math>[ABD] = \frac{2}{5}[ABC] = \frac{2}{5} \cdot 6 = \boxed{\textbf{(D) } \frac{12}{5}}</math> | ||
− | ==Solution 3== | + | ===Solution 3=== |
Since point <math>D</math> is on line <math>BC</math>, it will split it into <math>CD</math> and <math>DB</math>. Let <math>CD = 5 - x</math> and <math>DB = x</math>. Triangle <math>CAD</math> has side lengths <math>3, 5 - x, AD</math> and triangle <math>DAB</math> has side lengths <math>x, 4, AD</math>. Since both perimeters are equal, we have the equation <math>3 + 5 - x + AD = 4 + x + AD</math>. Eliminating <math>AD</math> and solving the resulting linear equation gives <math>x = 2</math>. Draw a perpendicular from point <math>D</math> to <math>AB</math>. Call the point of intersection <math>F</math>. Because angle <math>ABC</math> is common to both triangles <math>DBF</math> and <math>ABC</math>, and both are right triangles, both are similar. The hypotenuse of triangle <math>DBF</math> is 2, so the altitude must be <math>6/5</math> Because <math>DBF</math> and <math>ABD</math> share the same altitude, the height of <math>ABD</math> therefore must be <math>6/5</math>. The base of <math>ABD</math> is 4, so <math>[ABD] = \frac{1}{2} \cdot 4 \cdot \frac{6}{5} = \frac{12}{5} \implies \boxed{\textbf{(D) } \frac{12}{5}}</math> | Since point <math>D</math> is on line <math>BC</math>, it will split it into <math>CD</math> and <math>DB</math>. Let <math>CD = 5 - x</math> and <math>DB = x</math>. Triangle <math>CAD</math> has side lengths <math>3, 5 - x, AD</math> and triangle <math>DAB</math> has side lengths <math>x, 4, AD</math>. Since both perimeters are equal, we have the equation <math>3 + 5 - x + AD = 4 + x + AD</math>. Eliminating <math>AD</math> and solving the resulting linear equation gives <math>x = 2</math>. Draw a perpendicular from point <math>D</math> to <math>AB</math>. Call the point of intersection <math>F</math>. Because angle <math>ABC</math> is common to both triangles <math>DBF</math> and <math>ABC</math>, and both are right triangles, both are similar. The hypotenuse of triangle <math>DBF</math> is 2, so the altitude must be <math>6/5</math> Because <math>DBF</math> and <math>ABD</math> share the same altitude, the height of <math>ABD</math> therefore must be <math>6/5</math>. The base of <math>ABD</math> is 4, so <math>[ABD] = \frac{1}{2} \cdot 4 \cdot \frac{6}{5} = \frac{12}{5} \implies \boxed{\textbf{(D) } \frac{12}{5}}</math> | ||
− | ==Solution 4== | + | ===Solution 4=== |
Using any preferred method, realize <math>BD = 2</math>. Since we are given a 3-4-5 right triangle, we know the value of <math>\sin(\angle ABC) = \frac{3}{5}</math>. Since we are given <math>AB = 4</math>, apply the Sine Area Formula to get <math>\frac{1}{2} \cdot 4 \cdot 2 \cdot \frac{3}{5} = \boxed{\textbf{(D) } \frac{12}{5}}</math>. | Using any preferred method, realize <math>BD = 2</math>. Since we are given a 3-4-5 right triangle, we know the value of <math>\sin(\angle ABC) = \frac{3}{5}</math>. Since we are given <math>AB = 4</math>, apply the Sine Area Formula to get <math>\frac{1}{2} \cdot 4 \cdot 2 \cdot \frac{3}{5} = \boxed{\textbf{(D) } \frac{12}{5}}</math>. | ||
Revision as of 16:47, 16 January 2021
Contents
Problem 16
In the figure below, choose point on
so that
and
have equal perimeters. What is the area of
?
Solutions
Solution 1
Essentially, we see that if we draw a line from point A to imaginary point D, that line would apply to both triangles. Let us say that is the lenmeter of
would be
, while the perimeter of
would be
. Notice that we can find out
from these two equations. We can find out that
, and because the height of the triangles is the same, the ratio of the areas is
, so that means that the area of
.
Solution 2
We know that the perimeters of the two small triangles are and
. Setting both equal and using
, we have
and
. Now, we simply have to find the area of
. Since
, we must have
. Combining this with the fact that
, we get
Solution 3
Since point is on line
, it will split it into
and
. Let
and
. Triangle
has side lengths
and triangle
has side lengths
. Since both perimeters are equal, we have the equation
. Eliminating
and solving the resulting linear equation gives
. Draw a perpendicular from point
to
. Call the point of intersection
. Because angle
is common to both triangles
and
, and both are right triangles, both are similar. The hypotenuse of triangle
is 2, so the altitude must be
Because
and
share the same altitude, the height of
therefore must be
. The base of
is 4, so
Solution 4
Using any preferred method, realize . Since we are given a 3-4-5 right triangle, we know the value of
. Since we are given
, apply the Sine Area Formula to get
.
See Also
2017 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 | ||
All AJHSME/AMC 8 Problems and Solutions |
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