Difference between revisions of "2021 AMC 12A Problems/Problem 17"
MRENTHUSIASM (talk | contribs) m (→Solution 2 (One Pair of Similar Triangles, Then Areas): Varied the word choice other than "we have".) |
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<cmath>\begin{align*} | <cmath>\begin{align*} | ||
2[ADO]&=2[BCO] \\ | 2[ADO]&=2[BCO] \\ | ||
− | \frac{ | + | DO\cdot AD&=OB\cdot CP \\ |
− | x^2&= | + | x\left(\frac{hx}{11}\right)&=(x+22)h \\ |
− | (x | + | x^2&=11(x+22). |
− | |||
\end{align*}</cmath> | \end{align*}</cmath> | ||
+ | Rearranging and factoring result <math>(x-22)(x+11)=0,</math> from which <math>x=22.</math> | ||
− | + | Applying the Pythagorean Theorem on right <math>\triangle CPB,</math> we obtain <cmath>h=\sqrt{43^2-33^2}=\sqrt{(43+33)(43-33)}=\sqrt{760}=2\sqrt{190}.</cmath> | |
+ | |||
+ | Finally, we conclude that <math>AD=h\cdot\frac{x}{11}=4\sqrt{190},</math> so the answer is <math>4+190=\boxed{\textbf{(D) }194}.</math> | ||
~MRENTHUSIASM | ~MRENTHUSIASM |
Revision as of 19:48, 10 April 2021
- The following problem is from both the 2021 AMC 10A #17 and 2021 AMC 12A #17, so both problems redirect to this page.
Contents
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 (by Geometry Expressions)
Solution 1
Angle chasing reveals that , therefore Additional angle chasing shows that , therefore Since is right, the Pythagorean theorem implies that
~mn28407
Solution 2 (One Pair of Similar Triangles, Then Areas)
Since is isosceles with legs and it follows that the median is also an altitude of Let and We have
Since by AA, we get
Let the brackets denote areas. Notice that (By the same base/height, Subtracting from both sides gives ). Doubling both sides produces Rearranging and factoring result from which
Applying the Pythagorean Theorem on right we obtain
Finally, we conclude that so the answer is
~MRENTHUSIASM
Solution 3 (short)
Let and is perpendicular bisector of Let so
(1) so we get or
(2) pythag on gives
(3) with ratio so
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 , .
~ ihatemath123
Video Solution (Using Similar Triangles, Pythagorean Theorem)
~ pi_is_3.14
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=rtdovluzgQs
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.