Difference between revisions of "2021 AMC 12A Problems/Problem 17"
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(2) pythag on <math>\triangle CDP</math> gives <math>(11+x)^2 + y^2 = 43^2.</math> | (2) pythag on <math>\triangle CDP</math> gives <math>(11+x)^2 + y^2 = 43^2.</math> | ||
− | (3) <math>\triangle BPC \sim \triangle BDA</math> with ratio <math>1:2,</math> so <math>AD = 2y.</math> | + | (3) <math>\triangle BPC \sim \triangle BDA</math> with ratio <math>1:2,</math> so <math>AD = 2y.</math> (remember that <math>P</math> is the midpoint of <math>BD</math>) |
Thus, <math>xy/11 = 2y,</math> or <math>x = 22.</math> And <math>y = \sqrt{43^2 - 33^2} = 2 \sqrt{190},</math> so <math>AD = 4 \sqrt{190}</math> and the answer is <math>\boxed{194}.</math> | Thus, <math>xy/11 = 2y,</math> or <math>x = 22.</math> And <math>y = \sqrt{43^2 - 33^2} = 2 \sqrt{190},</math> so <math>AD = 4 \sqrt{190}</math> and the answer is <math>\boxed{194}.</math> |
Revision as of 23:50, 27 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
[hide]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 (Similar Triangles, Areas, Pythagorean Theorem)
Since is isosceles with legs
and
it follows that the median
is also an altitude. Let
and
We have
Since by vertical angles, we get
by AA, from which
or
Let the brackets denote areas. Notice that (By the same base and height,
Subtracting
from both sides gives
). Doubling both sides produces
Rearranging and factoring result
from which
Applying the Pythagorean Theorem to 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
(remember 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
,
.
~ 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.