Difference between revisions of "2021 AMC 12A Problems/Problem 17"
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==Diagram== | ==Diagram== | ||
− | + | <asy> | |
/* Made by MRENTHUSIASM */ | /* Made by MRENTHUSIASM */ | ||
size(300); | size(300); | ||
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O = intersectionpoint(A--C,B--D); | O = intersectionpoint(A--C,B--D); | ||
// R = intersectionpoints(Circle(O,7),Circle(X,3))[1]; | // R = intersectionpoints(Circle(O,7),Circle(X,3))[1]; | ||
− | dot(" | + | dot("$C$",C,1.5*SE,linewidth(4)); |
− | dot(" | + | dot("$D$",D,1.5*SW,linewidth(4)); |
− | dot(" | + | dot("$B$",B,1.5*NE,linewidth(4)); |
− | dot(" | + | dot("$A$",A,1.5*NW,linewidth(4)); |
− | dot(" | + | dot("$P$",P,1.5*N,linewidth(4)); |
− | dot(" | + | dot("$O$",O,1.5*S,linewidth(4)); |
markscalefactor=0.25; | markscalefactor=0.25; | ||
draw(rightanglemark(A,D,O),red); | draw(rightanglemark(A,D,O),red); | ||
draw(A--B--C--D--cycle^^A--C^^B--D^^C--P); | draw(A--B--C--D--cycle^^A--C^^B--D^^C--P); | ||
− | label(" | + | label("$43$",B--C,E); |
− | label(" | + | label("$43$",C--D,S); |
− | label(" | + | label("$11$",midpoint(O--P),NW); |
− | + | </asy> | |
~MRENTHUSIASM | ~MRENTHUSIASM | ||
Revision as of 10:18, 29 September 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
- 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 Bash)
- 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 Bash)
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
thus
Then by Pythagorean theorem on
(
is isosceles) we have
~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.