2021 AMC 12A Problems/Problem 17
- The following problem is from both the 2021 AMC 10A #17 and 2021 AMC 12A #17, so both problems redirect to this page.
Contents
- 1 Problem
- 2 Diagram
- 3 Solution 1 (Similar Triangles and Pythagorean Theorem)
- 4 Solution 2 (Similar Triangles, Areas, Pythagorean Theorem)
- 5 Solution 3 (Short)
- 6 Solution 4 (Extending the Line)
- 7 Solution 5
- 8 Solution 6 (Coordinate Geometry)
- 9 Video Solution (Using Similar Triangles, Pythagorean Theorem)
- 10 Video Solution by Punxsutawney Phil
- 11 Video Solution by Mathematical Dexterity
- 12 See also
Problem
Trapezoid has , and . Let be the intersection of the diagonals and , and let be the midpoint of . Given that , the length of can be written in the form , where and are positive integers and is not divisible by the square of any prime. What is ?
Diagram
~MRENTHUSIASM
Solution 1 (Similar Triangles and Pythagorean Theorem)
Angle chasing* reveals that , therefore or .
Additional angle chasing shows that , therefore or and .
Since is right, the Pythagorean theorem implies that The answer is .
- Angle Chasing: If we set , then we know that because is isosceles. And because , we conclude that too. Lastly, because and are both right triangles, they are similar by AA.
~mn28407 (Solution)
~mm (Angle Chasing Remark)
~eagleye ~MRENTHUSIASM (Minor Edits)
Solution 2 (Similar Triangles, Areas, Pythagorean Theorem)
Since is isosceles with base it follows that median is also an altitude. Let and so
Since by vertical angles, we conclude that by AA, from which or Let the brackets denote areas. Notice that (By the same base and height, we deduce that Subtracting from both sides gives ). Doubling both sides produces Rearranging and factoring result in from which
Applying the Pythagorean Theorem to right we have Finally, we get so the answer is
~MRENTHUSIASM
Solution 3 (Short)
Let . a is perpendicular bisector of Then, let thus
(1) so we get or
(2) Applying Pythagorean Theorem on gives
(3) with ratio so using the fact that is the midpoint of .
Thus, or And so and the answer is
~ ccx09
Solution 4 (Extending the Line)
Observe that is congruent to ; both are similar to . Let's extend and past points and respectively, such that they intersect at a point . Observe that is degrees, and that . Thus, by ASA, we know that , thus, , meaning is the midpoint of . Let be the midpoint of . Note that is congruent to , thus , meaning is the midpoint of
Therefore, and are both medians of . This means that is the centroid of ; therefore, because the centroid divides the median in a 2:1 ratio, . Recall that is the midpoint of ; . The question tells us that ; ; we can write this in terms of ; .
We are almost finished. Each side length of is twice as long as the corresponding side length or , since those triangles are similar; this means that . Now, by Pythagorean theorem on , .
The answer is .
~ ihatemath123
Solution 5
Since is the midpoint of isosceles triangle , it would be pretty easy to see that . Since as well, . Connecting , it’s obvious that . Since , .
Since is the midpoint of , the height of on side is half that of on . Since , .
As a basic property of a trapezoid, , so , or . Letting , then , and . Hence and .
Since , . Since , .
So, . The correct answer is
Solution 6 (Coordinate Geometry)
Let be the origin of the cartesian coordinate plane, lie on the positive -axis, and lie on the negative -axis. Then let the coordinates of Then the slope of is Since the slope of is the same. Note that as is isosceles lies on Thus since has equation ( is the origin), Therefore has equation and intersects (-axis) at The midpoint of is so from which Then by Pythagorean theorem on ( is isosceles), we have so
Finally, the answer is
~Aaryabhatta1
Video Solution (Using Similar Triangles, Pythagorean Theorem)
~ pi_is_3.14
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=rtdovluzgQs
Video Solution by Mathematical Dexterity
https://www.youtube.com/watch?v=QzAVdsgBBqg
See also
2021 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 |
2021 AMC 12A (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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.