Difference between revisions of "2012 AMC 12A Problems/Problem 12"
m (→Solution 3) |
m (→Solution 3) |
||
(One intermediate revision by the same user not shown) | |||
Line 63: | Line 63: | ||
Since CD passes through point (0,1), let C be (-x,1) and D be (x,1) | Since CD passes through point (0,1), let C be (-x,1) and D be (x,1) | ||
− | Point B is (-x, | + | Point B is <math>\left(-x, \sqrt{4-x^2}\right)</math> and Point A is <math>\left(x,\sqrt{4-x^2}\right)</math>. Now, we see that AB = 2x and BC = <math>\sqrt{4-x^2}</math>-1 |
Setting those two equal, we get the same equation as the previous solutions: <math>5x^2 +4x -3 = 0</math> and then solve the same way. | Setting those two equal, we get the same equation as the previous solutions: <math>5x^2 +4x -3 = 0</math> and then solve the same way. |
Latest revision as of 20:42, 3 January 2020
Problem
A square region is externally tangent to the circle with equation
at the point
on the side
. Vertices
and
are on the circle with equation
. What is the side length of this square?
Solution 1
![[asy] unitsize(15mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; real a=1; real b=2; pair O=(0,0); pair A=(-(sqrt(19)-2)/5,1); pair B=((sqrt(19)-2)/5,1); pair C=((sqrt(19)-2)/5,1+2(sqrt(19)-2)/5); pair D=(-(sqrt(19)-2)/5,1+2(sqrt(19)-2)/5); pair E=(-(sqrt(19)-2)/5,0); path inner=Circle(O,a); path outer=Circle(O,b); draw(outer); draw(inner); draw(A--B--C--D--cycle); draw(O--D--E--cycle); label("$A$",D,NW); label("$E$",E,SW); label("$O$",O,SE); label("$s+1$",(D--E),W); label("$\frac{s}{2}$",(E--O),S); pair[] ps={A,B,C,D,E,O}; dot(ps); [/asy]](http://latex.artofproblemsolving.com/1/7/b/17b54bd31c2906599eee882f7b6902279c53365e.png)
The circles have radii of and
. Draw the triangle shown in the figure above and write expressions in terms of
(length of the side of the square) for the sides of the triangle. Because
is the radius of the larger circle, which is equal to
, we can write the Pythagorean Theorem.
Use the quadratic formula.
Solution 2
Using the diagram above, we look at the top-right vertex of the square. Let us call this point . Then, we that since the square is symmetrical over the y-axis, that the y value is equal to
, since we can multiply the x value(which is half of
) by two to get
, and we add one since the square lies one unit above the origin. Now, all we must do is find the intersection of the larger circle,
, and the line
. Substituting the second equation into the first, we get:
Using the quadratic formula, we arrive with . However, recall that the x value is only one half of the side length. Multiplying this value by
, then, and using only the positive root(since the top right vertex of the square has a positive x value), we get:
Solution 3
Like the previous solutions, we can say that the square is symmetrical with respect to the y-axis.
Since CD passes through point (0,1), let C be (-x,1) and D be (x,1)
Point B is and Point A is
. Now, we see that AB = 2x and BC =
-1
Setting those two equal, we get the same equation as the previous solutions: and then solve the same way.
-Conantwiz2023
See Also
2012 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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.