Difference between revisions of "2015 AMC 10B Problems/Problem 19"
(→Solution 2) |
(→Solution 2) |
||
(18 intermediate revisions by 3 users not shown) | |||
Line 55: | Line 55: | ||
==Solution 2== | ==Solution 2== | ||
− | + | Let <math>AC = b</math> and <math>BC = a</math> (and we're given that <math>AB=12</math>). Draw line segments <math>YZ</math> and <math>WX</math>. Now we have cyclic quadrilateral <math>WXYZ.</math> | |
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
+ | This means that opposite angles sum to <math>180^{\circ}</math>. Therefore, <math>90 + m\angle YZA + 90 - m\angle WXB = 180</math>. Simplifying carefully, we get <math>m\angle YZA = m\angle WXB</math>. Similarly, <math>m\angle{ZYA}</math> = <math>m\angle{XWB}</math>. | ||
− | + | That means <math>\triangle ZYA \sim \triangle XWB</math>. | |
− | + | ||
− | + | Setting up proportions, | |
− | + | <math>\dfrac{b}{12}=\dfrac{12}{a+b}.</math> | |
− | + | Cross-multiplying we get: | |
− | + | <math>b^2+ab=12^2</math> | |
− | + | ||
− | + | But also, by Pythagoras, | |
− | + | <math>b^2+a^2=12^2</math>, so <math>ab=a^2 \Rightarrow a=b</math> | |
− | + | ||
− | + | Therefore, <math>\triangle ABC</math> is an isosceles right triangle. <math>AC=BC=\dfrac{12}{\sqrt{2}}=6\sqrt{2}</math>, so the perimeter is <cmath>\boxed{\textbf{(C)}\ 12+12\sqrt{2}}</cmath> | |
− | + | ||
− | + | ~BakedPotato66 | |
− | + | ||
− | + | ~LegionOfAvatars | |
− | + | ||
− | + | ==Solution 3 - Fakesolve== | |
− | + | Temporarily assume that we are as dumb as samrocksnature. Then, we would misread the instructions and draw the squares inside, not outside, the triangle. In addition, we would make a plethora of ridiculous assumptions. | |
− | + | ||
− | + | Our diagram would essentially be two squares intersecting in a square, such that a circle can be drawn through the two leftmost vertices of the left square and the two rightmost vertices of the right square. Since the diagonal of the square is given to be <math>12</math>, our answer is <math>12+6\sqrt{2}+6\sqrt{2} \Rightarrow \boxed{C}</math>. | |
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | </ | ||
==See Also== | ==See Also== |
Latest revision as of 01:34, 20 January 2021
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
The center of the circle lies on the perpendicular bisectors of both chords and . Therefore we know the center of the circle must also be the midpoint of the hypotenuse. Let this point be . Draw perpendiculars to and from , and connect and . . Let and . Then . Simplifying this gives . But by Pythagorean Theorem on , we know , because . Thus . So our equation simplifies further to . However , so , which means , or . Aha! This means is just an isosceles right triangle, so , and thus the perimeter is .
Solution 2
Let and (and we're given that ). Draw line segments and . Now we have cyclic quadrilateral
This means that opposite angles sum to . Therefore, . Simplifying carefully, we get . Similarly, = .
That means .
Setting up proportions, Cross-multiplying we get:
But also, by Pythagoras, , so
Therefore, is an isosceles right triangle. , so the perimeter is
~BakedPotato66
~LegionOfAvatars
Solution 3 - Fakesolve
Temporarily assume that we are as dumb as samrocksnature. Then, we would misread the instructions and draw the squares inside, not outside, the triangle. In addition, we would make a plethora of ridiculous assumptions.
Our diagram would essentially be two squares intersecting in a square, such that a circle can be drawn through the two leftmost vertices of the left square and the two rightmost vertices of the right square. Since the diagonal of the square is given to be , our answer is .
See Also
2015 AMC 10B (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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.