Difference between revisions of "2022 AIME I Problems/Problem 11"
(→Solution 1 (No trig)) |
(→Solution 1 (No trig)) |
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defaultpen(linewidth(0.6)+fontsize(11)); | defaultpen(linewidth(0.6)+fontsize(11)); | ||
size(20cm); | size(20cm); | ||
− | pair A,B,C,D,P,Q,O; | + | pair A,B,C,D,E,F,P,Q,O; |
A=(0,0); | A=(0,0); | ||
+ | E = (24,15); | ||
+ | F = (30,0); | ||
O = (10.5,7.5); | O = (10.5,7.5); | ||
label("$A$", A, SW); | label("$A$", A, SW); | ||
Line 55: | Line 57: | ||
draw(C--(30,0),dashed); | draw(C--(30,0),dashed); | ||
draw(D--(30,0)); | draw(D--(30,0)); | ||
+ | dot(E); | ||
+ | dot(F); | ||
label("$3$", midpoint(A--P), S); | label("$3$", midpoint(A--P), S); | ||
Line 71: | Line 75: | ||
label("$T_2$", (10.5,0), S); | label("$T_2$", (10.5,0), S); | ||
label("$T_3$", (4.5,11.25),W); | label("$T_3$", (4.5,11.25),W); | ||
+ | label("$E$",E, N); | ||
+ | label("$F$",F, S); | ||
+ | |||
+ | |||
</asy> | </asy> | ||
Line 76: | Line 84: | ||
We obviously see that we must use power of a point since they've given us lengths in a circle and there are intersection points. Let <math>T_1, T_2, T_3</math> be our tangents from the circle to the parallelogram. By the secant power of a point, the power of <math>A = 3 \cdot (3+9) = 36</math>. Then <math>AT_2 = AT_3 = \sqrt{36} = 6</math>. Similarly, the power of <math>C = 16 \cdot (16+9) = 400</math> and <math>CT_1 = \sqrt{400} = 20</math>. We let <math>BT_3 = BT_1 = x</math> and label the diagram accordingly. | We obviously see that we must use power of a point since they've given us lengths in a circle and there are intersection points. Let <math>T_1, T_2, T_3</math> be our tangents from the circle to the parallelogram. By the secant power of a point, the power of <math>A = 3 \cdot (3+9) = 36</math>. Then <math>AT_2 = AT_3 = \sqrt{36} = 6</math>. Similarly, the power of <math>C = 16 \cdot (16+9) = 400</math> and <math>CT_1 = \sqrt{400} = 20</math>. We let <math>BT_3 = BT_1 = x</math> and label the diagram accordingly. | ||
− | Notice that because <math>BC = AD, 20+x = 6+DT_2 \implies DT_2 = 14+x</math>. Let <math>O</math> be the center of the circle. Since <math>OT_1</math> and <math>OT_2</math> intersect <math>BC</math> and <math>AD</math>, respectively, at right angles, we have <math> | + | Notice that because <math>BC = AD, 20+x = 6+DT_2 \implies DT_2 = 14+x</math>. Let <math>O</math> be the center of the circle. Since <math>OT_1</math> and <math>OT_2</math> intersect <math>BC</math> and <math>AD</math>, respectively, at right angles, we have <math>T_2T_2CD</math> is a right-angled trapezoid and more importantly, the diameter of the circle is the height of the triangle. Therefore, we can drop an altitude from <math>D</math> to <math>BC</math> and <math>C</math> to <math>AD</math>, and both are equal to <math>2r</math>. Since <math>T_1E = T_2D</math>, <math>20 - CE = 14+x \implies CE = 6-x</math>. Since <math>CE = DF, DF = 6-x</math> and <math>AF = 6+14+x+6-x = 26</math>. We can now use Pythagorean theorem on <math>\triangle ACF</math>; we have <math>26^2 + (2r)^2 = (3+9+16)^2 \implies 4r^2 = 784-676 \implies 4r^2 = 108 \implies 2r = 6\sqrt{3}</math> and <math>r^2 = 27</math>. |
− | |||
+ | We know that <math>CD = 6+x</math> because <math>ABCD</math> is a parallelogram. Using Pythagorean theorem on <math>\triangle CDF</math>, <math>(6+x)^2 = (6-x)^2 + 108 \implies (6+x)^2-(6-x)^2 = 108 \implies 12 \cdot 2x = 108 \implies 2x = 9 \implies x = \frac{9}{2}</math>. Therefore, base <math>BC = 20 + \frac{9}{2} = \frac{49}{2}</math>. Thus the area of the parallelogram is the base times the height, which is <math>\frac{49}{2} \cdot 6\sqrt{3} = 147\sqrt{3}</math> and the answer is <math>\boxed{150}</math> | ||
− | |||
~KingRavi | ~KingRavi |
Revision as of 16:30, 21 February 2022
Contents
Problem
Let be a parallelogram with
. A circle tangent to sides
,
, and
intersects diagonal
at points
and
with
, as shown. Suppose that
,
, and
. Then the area of
can be expressed in the form
, where
and
are positive integers, and
is not divisible by the square of any prime. Find
.
Solution 1 (No trig)
Let's redraw the diagram, but extend some helpful lines.
We obviously see that we must use power of a point since they've given us lengths in a circle and there are intersection points. Let be our tangents from the circle to the parallelogram. By the secant power of a point, the power of
. Then
. Similarly, the power of
and
. We let
and label the diagram accordingly.
Notice that because . Let
be the center of the circle. Since
and
intersect
and
, respectively, at right angles, we have
is a right-angled trapezoid and more importantly, the diameter of the circle is the height of the triangle. Therefore, we can drop an altitude from
to
and
to
, and both are equal to
. Since
,
. Since
and
. We can now use Pythagorean theorem on
; we have
and
.
We know that because
is a parallelogram. Using Pythagorean theorem on
,
. Therefore, base
. Thus the area of the parallelogram is the base times the height, which is
and the answer is
~KingRavi
Solution 2
Let the circle tangent to at
separately, denote that
Using POP, it is very clear that , let
, using LOC in
,
, similarly, use LOC in
, getting that
. We use the second equation to minus the first equation, getting that
, we can get
.
Now applying LOC in , getting
, solving this equation to get
, then
,
, the area is
leads to
~bluesoul
Solution 3
Denote by the center of the circle. Denote by
the radius of the circle.
Denote by
,
,
the points that the circle meets
,
,
at, respectively.
Because the circle is tangent to ,
,
,
,
,
,
.
Because ,
,
,
are collinear.
Following from the power of a point, . Hence,
.
Following from the power of a point, . Hence,
.
Denote . Because
and
are tangents to the circle,
.
Because is a right trapezoid,
.
Hence,
.
This can be simplified as
\[
6 x = r^2 . \hspace{1cm} (1)
\]
In , by applying the law of cosines, we have
\begin{align*}
AC^2 & = AB^2 + CB^2 - 2 AB \cdot CB \cos B \\
& = AB^2 + CB^2 + 2 AB \cdot CB \cos A \\
& = AB^2 + CB^2 + 2 AB \cdot CB \cdot \frac{AE - BF}{AB} \\
& = AB^2 + CB^2 + 2 CB \left( AE - BF \right) \\
& = \left( 6 + x \right)^2 + \left( 20 + x \right)^2 + 2 \left( 20 + x \right) \left( 6 - x \right) \\
& = 24 x + 676 .
\end{align*}
Because , we get
.
Plugging this into Equation (1), we get
.
Therefore, \begin{align*} {\rm Area} \ ABCD & = CB \cdot EF \\ & = \left( 20 + x \right) \cdot 2r \\ & = 147 \sqrt{3} . \end{align*}
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
Solution 4
Let be the circle, let
be the radius of
, and let the points at which
is tangent to
,
, and
be
,
, and
, respectively. Note that PoP on
and
with respect to
yields
and
. We can compute the area of
in two ways:
1. By the half-base-height formula, .
2. We can drop altitudes from the center of
to
,
, and
, which have lengths
,
, and
. Thus,
.
Equating the two expressions for and solving for
yields
.
Let . By the Parallelogram Law,
. Solving for
yields
. Thus,
, for a final answer of
.
~ Leo.Euler
Video Solution
https://www.youtube.com/watch?v=FeM_xXiJj0c&t=1s
~Steven Chen (www.professorchenedu.com)
Video Solution 2 (Mathematical Dexterity)
https://www.youtube.com/watch?v=1nDKQkr9NaU
See Also
2022 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.