Difference between revisions of "2017 AMC 8 Problems/Problem 16"
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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 | + | ==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*4}{2} = 6</math>, we get <math>[ABD] = \frac{2}{5}[ABC] = \frac{2}{5} * 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*4}{2} = 6</math>, we get <math>[ABD] = \frac{2}{5}[ABC] = \frac{2}{5} * 6 = \boxed{\textbf{(D) } \frac{12}{5}}</math> | ||
Revision as of 17:10, 22 November 2017
Contents
Problem 16
In the figure below, choose point 
on 
so that 
and 
have equal perimeters. What is the area of ?
![[asy]draw((0,0)--(4,0)--(0,3)--(0,0)); label("$A$", (0,0), SW); label("$B$", (4,0), ESE); label("$C$", (0, 3), N); label("$3$", (0, 1.5), W); label("$4$", (2, 0), S); label("$5$", (2, 1.5), NE);[/asy]](http://latex.artofproblemsolving.com/5/3/e/53eb67d35ecca7ac6f68e8dff60b0eb5f3e4a602.png)
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 ![$\frac{[ABD]}{ACD]} = 2/3$](http://latex.artofproblemsolving.com/f/e/9/fe9792dcd5ec9e17365c91523549220808939768.png)
. Combining this with the fact that ![$[ABC] = [ABD] + [ACD] = \frac{3*4}{2} = 6$](http://latex.artofproblemsolving.com/6/4/e/64e2d85b6138ea15949b7362bd0d9b2204e8a359.png)
, we get ![$[ABD] = \frac{2}{5}[ABC] = \frac{2}{5} * 6 = \boxed{\textbf{(D) } \frac{12}{5}}$](http://latex.artofproblemsolving.com/5/8/3/583f355c81572d0b7d5075d6e2e3a9726fed0f35.png)
Solution 2
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 ![$[ABD] = \frac{1}{2} * 4 * \frac{6}{5} = \frac{12}{5} \implies \boxed{\textbf{(D) } \frac{12}{5}}$](http://latex.artofproblemsolving.com/4/4/b/44b6295bba45043902b514adb50a4b5039ab4b07.png)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
