Difference between revisions of "2023 AMC 10A Problems/Problem 17"
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==Solution== | ==Solution== | ||
| + | [insert asy diagram] | ||
| + | |||
| + | Using knowledge of common Pythagorean triples and guess and check, we can find that <math>\triangle{ABP}</math> is a <math>8</math>-<math>15</math>-<math>17</math> triangle with side lengths <math>16</math>-<math>30</math>-<math>34</math>, and <math>\triangle{PCQ}</math> and <math>\triangle{QDA}</math> are <math>3</math>-<math>4</math>-<math>5</math> triangles with side lengths <math>9</math>-<math>12</math>-<math>15</math> and <math>21</math>-<math>28</math>-<math>35</math>, respectively. | ||
| + | |||
| + | Adding up the side lengths of <math>\triangle{APQ}</math> gives <math>34+15+35=\boxed{\textbf{(A) } 84}.</math> | ||
| + | |||
| + | ~ItsMeNoobieboy | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2023|ab=A|num-b=16|num-a=18}} | {{AMC10 box|year=2023|ab=A|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 20:41, 9 November 2023
Let 
be a rectangle with 
and 
. Point 
and 
lie on 
and 
respectively so that all sides of 
and 
have integer lengths. What is the perimeter of 
?
Solution
[insert asy diagram]
Using knowledge of common Pythagorean triples and guess and check, we can find that 
is a 
-
-
triangle with side lengths 
-
-
, and 
and are 
-
-
triangles with side lengths 
-
- and 
-
-
, respectively.
Adding up the side lengths of gives 
~ItsMeNoobieboy
See Also
| 2023 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 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. 
