Difference between revisions of "2022 AMC 12B Problems/Problem 14"
MRENTHUSIASM (talk | contribs) (→Solution 5) |
MRENTHUSIASM (talk | contribs) (→Solution 1 (Dot Product)) |
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\textbf{(D)}\ \frac{1}{2} \qquad | \textbf{(D)}\ \frac{1}{2} \qquad | ||
\textbf{(E)}\ \frac{4}{7} \qquad</math> | \textbf{(E)}\ \frac{4}{7} \qquad</math> | ||
+ | |||
+ | ==Diagram== | ||
+ | <asy> | ||
+ | /* Made by MRENTHUSIASM */ | ||
+ | size(250); | ||
+ | |||
+ | real xMin = -15; | ||
+ | real xMax = 15; | ||
+ | real yMin = -17; | ||
+ | real yMax = 17; | ||
+ | |||
+ | draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); | ||
+ | draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); | ||
+ | label("$x$",(xMax,0),(2,0)); | ||
+ | label("$y$",(0,yMax),(0,2)); | ||
+ | |||
+ | real f(real x) { return x^2+2*x-15; } | ||
+ | draw(graph(f,-6.75,4.75),red); | ||
+ | |||
+ | pair A, B, C; | ||
+ | A = (-5,0); | ||
+ | B = (0,-15); | ||
+ | C = (3,0); | ||
+ | |||
+ | draw(A--B--C); | ||
+ | dot("$A$",A,1.5NW,linewidth(4.5)); | ||
+ | dot("$B$",B,1.5SE,linewidth(4.5)); | ||
+ | dot("$C$",C,1.5NE,linewidth(4.5)); | ||
+ | </asy> | ||
+ | ~MRENTHUSIASM | ||
== Solution 1 (Dot Product) == | == Solution 1 (Dot Product) == |
Revision as of 20:07, 9 January 2023
Contents
Problem
The graph of intersects the
-axis at points
and
and the
-axis at point
. What is
?
Diagram
~MRENTHUSIASM
Solution 1 (Dot Product)
First, find ,
, and
. Create vectors
and
These can be reduced to
and
, respectively. Then, we can use the dot product to calculate the cosine of the angle (where
) between them:
Thus,
~Indiiiigo
Solution 2
intersects the
-axis at points
and
. Without loss of generality, let these points be
and
respectively. Also, the graph intersects the y-axis at point
.
Let point denote the origin
. Note that triangles
and
are right.
We have
Alternatively, we can use the Pythagorean Theorem to find that and
and then use the
area formula for a triangle and the Law of Cosines to find
.
Solution 3
Like above, we set to
,
to
, and
to
, then finding via the Pythagorean Theorem that
and
. Using the Law of Cosines, we see that
Then, we use the identity
to get
~ jamesl123456
Solution 4
We can reflect the figure, but still have the same angle. This problem is the same as having points ,
, and
, where we're solving for angle FED. We can use the formula for
to solve now where
is the
-axis to angle
and
is the
-axis to angle
.
and
. Plugging these values into the
formula, we get
which is
~mathboy100 (minor LaTeX edits)
Solution 5
We use the identity
Note that has side-lengths
and
from Pythagorean theorem, with the area being
We equate the areas together to get:
from which
From Pythagorean identity,
Then we use , to obtain
- SAHANWIJETUNGA
See Also
2022 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
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.