Difference between revisions of "2018 AMC 10A Problems/Problem 13"
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<math>\textbf{(A) } 1+\frac12 \sqrt2 \qquad \textbf{(B) } \sqrt3 \qquad \textbf{(C) } \frac74 \qquad \textbf{(D) } \frac{15}{8} \qquad \textbf{(E) } 2 </math> | <math>\textbf{(A) } 1+\frac12 \sqrt2 \qquad \textbf{(B) } \sqrt3 \qquad \textbf{(C) } \frac74 \qquad \textbf{(D) } \frac{15}{8} \qquad \textbf{(E) } 2 </math> | ||
| − | ==Solution | + | ==Solution== |
First, we need to realize that the crease line is just the perpendicular bisector of side <math>AB</math>, the hypotenuse of right triangle <math>\triangle ABC</math>. Call the midpoint of <math>AC</math> <math>D</math>. Draw this line and call the intersection point with <math>AC</math> as <math>E</math>. Now, <math>\triangle ABC</math> is similar to <math>\triangle ADE</math> by <math>AA</math> similarity. Setting up the ratios, we find that | First, we need to realize that the crease line is just the perpendicular bisector of side <math>AB</math>, the hypotenuse of right triangle <math>\triangle ABC</math>. Call the midpoint of <math>AC</math> <math>D</math>. Draw this line and call the intersection point with <math>AC</math> as <math>E</math>. Now, <math>\triangle ABC</math> is similar to <math>\triangle ADE</math> by <math>AA</math> similarity. Setting up the ratios, we find that | ||
<cmath>\frac{BC}{AC}=\frac{DE}{AD} \Rightarrow \frac{3}{4}=\frac{4}{\frac{5}{2}} \Rightarrow DE=\frac{15}{8}.</cmath> | <cmath>\frac{BC}{AC}=\frac{DE}{AD} \Rightarrow \frac{3}{4}=\frac{4}{\frac{5}{2}} \Rightarrow DE=\frac{15}{8}.</cmath> | ||
Thus, our answer is choice <math>\boxed{D}</math>. | Thus, our answer is choice <math>\boxed{D}</math>. | ||
Revision as of 14:32, 8 February 2018
A paper triangle with sides of lengths 3,4, and 5 inches, as shon, is folded so that point 
falls on point 
. What is the length in inches of the crease?
![[asy] draw((0,0)--(4,0)--(4,3)--(0,0)); label("$A$", (0,0), SW); label("$B$", (4,3), NE); label("$C$", (4,0), SE); label("$4$", (2,0), S); label("$3$", (4,1.5), E); label("$5$", (2,1.5), NW); fill(origin--(0,0)--(4,3)--(4,0)--cycle, gray); [/asy]](http://latex.artofproblemsolving.com/2/1/1/21145c9f39a3f3cd34317f8b1f8e3f3541c3c477.png)
Solution
First, we need to realize that the crease line is just the perpendicular bisector of side 
, the hypotenuse of right triangle 
. Call the midpoint of 
. Draw this line and call the intersection point with as 
. Now, is similar to 
by 
similarity. Setting up the ratios, we find that
![\[\frac{BC}{AC}=\frac{DE}{AD} \Rightarrow \frac{3}{4}=\frac{4}{\frac{5}{2}} \Rightarrow DE=\frac{15}{8}.\]](http://latex.artofproblemsolving.com/e/6/5/e65eff09fa51ccdfe0f3fdf8e3e2c1948de6518c.png)
Thus, our answer is choice 
.
