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 1== |
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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> point <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> point <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 | ||
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~Nivek | ~Nivek | ||
− | Solution 2 (if you are already out of time) | + | ==Solution 2 (if you are already out of time)== |
Simply make a 3x4x5 inch triangle and then cut it out (using fine rips). Then, make the fold and mesure. It will be <math>\boxed{D} \frac{15}{8}</math> inches in length. | Simply make a 3x4x5 inch triangle and then cut it out (using fine rips). Then, make the fold and mesure. It will be <math>\boxed{D} \frac{15}{8}</math> inches in length. |
Revision as of 16:59, 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?
Solution 1
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 point . Draw this line and call the intersection point with as . Now, is similar to by similarity. Setting up the ratios, we find that Thus, our answer is .
~Nivek
Solution 2 (if you are already out of time)
Simply make a 3x4x5 inch triangle and then cut it out (using fine rips). Then, make the fold and mesure. It will be inches in length.
See Also
2018 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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 AMC 10 Problems and Solutions |
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