Difference between revisions of "2010 AIME II Problems/Problem 14"
(Added another solution to 2010 AIME II Problem #14.) |
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Let <math>\alpha=\angle{ACP}</math>, <math>\beta=\angle{ABC}</math>, and <math>x=BP</math>. By Law of Sines, | Let <math>\alpha=\angle{ACP}</math>, <math>\beta=\angle{ABC}</math>, and <math>x=BP</math>. By Law of Sines, | ||
− | <math>\frac{1}{sin(\beta)}=\frac{x}{sin(90-\alpha)}\implies sin(\beta)=\frac{cos(\alpha)}{x}</math> (1), and | + | <math>\frac{1}{\sin(\beta)}=\frac{x}{\sin(90-\alpha)}\implies \sin(\beta)=\frac{\cos(\alpha)}{x}</math> (1), and |
− | <math>\frac{4-x}{sin(\alpha)}=\frac{ | + | <math>\frac{4-x}{\sin(\alpha)}=\frac{4\sin(\beta)}{\sin(2\alpha)} \implies 4-x=\frac{2\sin(\beta)}{\cos(\alpha)}</math>. (2) |
Then, substituting (1) into (2), we get | Then, substituting (1) into (2), we get |
Revision as of 19:32, 2 June 2023
Contents
Problem
Triangle with right angle at
,
and
. Point
on
is chosen such that
and
. The ratio
can be represented in the form
, where
,
,
are positive integers and
is not divisible by the square of any prime. Find
.
Solution 1
Let be the circumcenter of
and let the intersection of
with the circumcircle be
. It now follows that
. Hence
is isosceles and
.
Denote the projection of
onto
. Now
. By the Pythagorean Theorem,
. Now note that
. By the Pythagorean Theorem,
. Hence it now follows that,
This gives that the answer is .
An alternate finish for this problem would be to use Power of a Point on and
. By Power of a Point Theorem,
. Since
, we can solve for
and
, giving the same values and answers as above.
![[asy] /* geogebra conversion, see azjps userscripts.org/scripts/show/72997 */ import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(250); real lsf = 0.5; /* changes label-to-point distance */ pen xdxdff = rgb(0.49,0.49,1); pen qqwuqq = rgb(0,0.39,0); pen fftttt = rgb(1,0.2,0.2); /* segments and figures */ draw((0.2,0.81)--(0.33,0.78)--(0.36,0.9)--(0.23,0.94)--cycle,qqwuqq); draw((0.81,-0.59)--(0.93,-0.54)--(0.89,-0.42)--(0.76,-0.47)--cycle,qqwuqq); draw(circle((2,0),2)); draw((0,0)--(0.23,0.94),linewidth(1.6pt)); draw((0.23,0.94)--(4,0),linewidth(1.6pt)); draw((0,0)--(4,0),linewidth(1.6pt)); draw((0.23,(+0.55-0.94*0.23)/0.35)--(4.67,(+0.55-0.94*4.67)/0.35)); /* points and labels */ label("$1$", (0.26,0.42), SE*lsf); draw((1.29,-1.87)--(2,0)); label("$2$", (2.91,-0.11), SE*lsf); label("$2$", (1.78,-0.82), SE*lsf); pair parametricplot0_cus(real t){ return (0.28*cos(t)+0.23,0.28*sin(t)+0.94); } draw(graph(parametricplot0_cus,-1.209429202888189,-0.24334747753738661)--(0.23,0.94)--cycle,fftttt); pair parametricplot1_cus(real t){ return (0.28*cos(t)+0.59,0.28*sin(t)+0); } draw(graph(parametricplot1_cus,0.0,1.9321634507016043)--(0.59,0)--cycle,fftttt); label("$\theta$", (0.42,0.77), SE*lsf); label("$2\theta$", (0.88,0.38), SE*lsf); draw((2,0)--(0.76,-0.47)); pair parametricplot2_cus(real t){ return (0.28*cos(t)+2,0.28*sin(t)+0); } draw(graph(parametricplot2_cus,-1.9321634507016048,0.0)--(2,0)--cycle,fftttt); label("$2\theta$", (2.18,-0.3), SE*lsf); dot((0,0)); label("$B$", (-0.21,-0.2),NE*lsf); dot((4,0)); label("$A$", (4.03,0.06),NE*lsf); dot((2,0)); label("$O$", (2.04,0.06),NE*lsf); dot((0.59,0)); label("$P$", (0.28,-0.27),NE*lsf); dot((0.23,0.94)); label("$C$", (0.07,1.02),NE*lsf); dot((1.29,-1.87)); label("$D$", (1.03,-2.12),NE*lsf); dot((0.76,-0.47)); label("$E$", (0.56,-0.79),NE*lsf); clip((-0.92,-2.46)--(-0.92,2.26)--(4.67,2.26)--(4.67,-2.46)--cycle); [/asy]](http://latex.artofproblemsolving.com/d/7/e/d7e4141491260a90ef8083f3335e1bda58389bce.png)
Solution 2
Let ,
by convention. Also, Let
and
. Finally, let
and
.
We are then looking for
Now, by arc interceptions and angle chasing we find that , and that therefore
Then, since
(it intercepts the same arc as
) and
is right,
.
Using law of sines on , we additionally find that
Simplification by the double angle formula
yields
.
We equate these expressions for to find that
. Since
, we have enough information to solve for
and
. We obtain
Since we know , we use
Solution 3
Let be equal to
. Then by Law of Sines,
and
. We then obtain
and
. Solving, we determine that
. Plugging this in gives that
. The answer is
.
Solution 4 (The quickest and most elegant)
Let ,
, and
. By Law of Sines,
(1), and
. (2)
Then, substituting (1) into (2), we get
The answer is .
~Rowechen
Solution 5
Let . Then,
and
. Let the foot of the angle bisector of
on side
be
. Then,
and
due to the angles of these triangles.
Let . By the Angle Bisector Theorem,
, so
. Moreover, since
, by similar triangle ratios,
. Therefore,
.
Construct the perpendicular from to
and denote it as
. Denote the midpoint of
as
. Since
is an angle bisector,
is congruent to
, so
.
Also, . Thus,
. After some major cancellation, we have
, which is a quadratic in
. Thus,
.
Taking the negative root implies , contradiction. Thus, we take the positive root to find that
. Thus,
, and our desired ratio is
.
The answer is .
Solution 6
Let be the circumcenter of
. Since
is a right triangle,
will be on
and
. Let
.
Next, let's do some angle chasing. Label , and
. Thus,
, and by isosceles triangles,
. Then, by angle subtraction,
.
Using the Law of Sines: Using trigonometric identies,
. Plugging this back into the Law of Sines formula gives us:
Next, using the Law of Cosines:
Substituting
gives us:
Solving for x gives
Finally: , which gives us an answer of
. ~adyj
See also
2010 AIME II (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 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.