Difference between revisions of "2015 AMC 12B Problems/Problem 19"
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− | The center of the circle on which <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> lie must be equidistant from each of these four points. Draw the perpendicular bisectors of <math>\overline{XY}</math> and of <math>\overline{WZ}</math>. Note that the perpendicular bisector of <math>\overline{ | + | The center of the circle on which <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> lie must be equidistant from each of these four points. Draw the perpendicular bisectors of <math>\overline{XY}</math> and of <math>\overline{WZ}</math>. Note that the perpendicular bisector of <math>\overline{WZ}</math> is parallel to <math>\overline{CW}</math> and passes through the midpoint of <math>\overline{AC}</math>. Therefore, the triangle that is formed by <math>A</math>, the midpoint of <math>\overline{AC}</math>, and the point at which this perpendicular bisector intersects <math>\overline{AB}</math> must be similar to <math>\triangle ABC</math>, and the ratio of a side of the smaller triangle to a side of <math>\triangle ABC</math> is 1:2. Consequently, the perpendicular bisector of <math>\overline{XY}</math> passes through the midpoint of <math>\overline{AB}</math>. The perpendicular bisector of <math>\overline{WZ}</math> must include the midpoint of <math>\overline{AB}</math> as well. Since all points on a perpendicular bisector of any two points <math>M</math> and <math>N</math> are equidistant from <math>M</math> and <math>N</math>, the center of the circle must be the midpoint of <math>\overline{AB}</math>. |
Now the distance between the midpoint of <math>\overline{AB}</math> and <math>Z</math>, which is equal to the radius of this circle, is <math>\sqrt{12^2 + 6^2} = \sqrt{180}</math>. Let <math>a=AC</math>. Then the distance between the midpoint of <math>\overline{AB}</math> and <math>Y</math>, also equal to the radius of the circle, is given by <math>\sqrt{\left(\frac{a}{2}\right)^2 + \left(a + \frac{\sqrt{144 - a^2}}{2}\right)^2}</math> (the ratio of the similar triangles is involved here). Squaring these two expressions for the radius and equating the results, we have | Now the distance between the midpoint of <math>\overline{AB}</math> and <math>Z</math>, which is equal to the radius of this circle, is <math>\sqrt{12^2 + 6^2} = \sqrt{180}</math>. Let <math>a=AC</math>. Then the distance between the midpoint of <math>\overline{AB}</math> and <math>Y</math>, also equal to the radius of the circle, is given by <math>\sqrt{\left(\frac{a}{2}\right)^2 + \left(a + \frac{\sqrt{144 - a^2}}{2}\right)^2}</math> (the ratio of the similar triangles is involved here). Squaring these two expressions for the radius and equating the results, we have |
Revision as of 12:40, 6 February 2017
Problem
In , and . Squares and are constructed outside of the triangle. The points , , , and lie on a circle. What is the perimeter of the triangle?
Solution 1
First, we should find the center and radius of this circle. We can find the center by drawing the perpendicular bisectors of and and finding their intersection point. This point happens to be the midpoint of , the hypotenuse. Let this point be . To find the radius, determine , where , , and . Thus, the radius .
Next we let and . Consider the right triangle first. Using the pythagorean theorem, we find that . Next, we let to be the midpoint of , and we consider right triangle . By the pythagorean theorem, we have that . Expanding this equation, we get that
This means that is a 45-45-90 triangle, so . Thus the perimeter is which is answer . image needed
Solution 2
The center of the circle on which , , , and lie must be equidistant from each of these four points. Draw the perpendicular bisectors of and of . Note that the perpendicular bisector of is parallel to and passes through the midpoint of . Therefore, the triangle that is formed by , the midpoint of , and the point at which this perpendicular bisector intersects must be similar to , and the ratio of a side of the smaller triangle to a side of is 1:2. Consequently, the perpendicular bisector of passes through the midpoint of . The perpendicular bisector of must include the midpoint of as well. Since all points on a perpendicular bisector of any two points and are equidistant from and , the center of the circle must be the midpoint of .
Now the distance between the midpoint of and , which is equal to the radius of this circle, is . Let . Then the distance between the midpoint of and , also equal to the radius of the circle, is given by (the ratio of the similar triangles is involved here). Squaring these two expressions for the radius and equating the results, we have
Since cannot be equal to 12, the length of the hypotenuse of the right triangle, we can divide by , and arrive at . The length of other leg of the triangle must be . Thus, the perimeter of the triangle is .
Solution 3
In order to solve this problem, we can search for similar triangles. Begin by drawing triangle and squares and . Draw segments and . Because we are given points , , , and lie on a circle, we can conclude that forms a cyclic quadrilateral. Take and extend it through a point on . Now, we must do some angle chasing to prove that is similar to .
Let denote the measure of . Following this, measures . By our construction, is a straight line, and we know is a right angle. Therefore, measures . Also, is a right angle and thus, is a right angle. Sum and to find , which measures . We also know that measures . Therefore, .
Let denote the measure of . It follows that measures . Because is a cyclic quadrilateral, . Therefore, must measure , and must measure . Therefore, .
and , so ! Let . By Pythagorean theorem, . Now we have , , , and . We can set up an equation:
Solving for , we find that or , which we omit. The perimeter of the triangle is . Plugging in , we get .
See Also
2015 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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 12 Problems and Solutions |
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