Difference between revisions of "1983 AIME Problems/Problem 14"
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Thus by Power of a Point in the circle passing through <math>Q</math>, <math>R</math>, and <math>S</math>, we have <math>x \cdot 2x = 10 \cdot (10+16) = 260</math>, so <math>x^2 = \boxed{130}</math>. | Thus by Power of a Point in the circle passing through <math>Q</math>, <math>R</math>, and <math>S</math>, we have <math>x \cdot 2x = 10 \cdot (10+16) = 260</math>, so <math>x^2 = \boxed{130}</math>. | ||
+ | |||
+ | ===Full Proof that R, A, B are collinear === | ||
+ | <asy> | ||
+ | size(0,5cm); | ||
+ | pair a=(8,0),b=(20,0),t=(14,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); | ||
+ | draw(b--r--n--b--a--m--n); | ||
+ | draw(a--q--m); | ||
+ | draw(circumcircle(origin,q,p)); | ||
+ | draw(circumcircle((14,0),p,r)); | ||
+ | draw(rightanglemark(a,m,n,24)); | ||
+ | draw(rightanglemark(b,n,r,24)); | ||
+ | label("$A$",a,S); | ||
+ | label("$B$",b,S); | ||
+ | label("$M$",m,NE); | ||
+ | label("$N$",n,NE); | ||
+ | label("$P$",p,N); | ||
+ | label("$Q$",q,NW); | ||
+ | label("$R$",r,E); | ||
+ | label("$12$",(14,0),SW); | ||
+ | label("$6$",(23,0),S); | ||
+ | label("$T$", t , NW); | ||
+ | </asy> | ||
+ | |||
+ | Let <math>M</math> and <math>N</math> be the feet of the perpendicular from <math>A</math> to <math>PQ</math> and <math>B</math> to <math>PR</math> respectively. It is well known that a perpendicular from the center of a circle to a chord of that circle bisects the chord, so <math>QM = MP = PN = NR</math>, since the problem told us <math>QP = PR</math>. | ||
+ | |||
+ | We will show that <math>R</math> lies on <math>AB</math>. | ||
+ | |||
+ | Let <math>T</math> be the intersection of circle centered at <math>B</math> with <math>AB</math>. Then <math>BT = TA = 6</math>. | ||
+ | |||
+ | Let <math>P</math>' be the foot of the perpendicular from <math>T</math> to <math>MN</math>. Then <math>TP'</math> is a midline (or midsegment) in trapezoid <math>AMNB</math>, so <math>P'</math> coincides with <math>P</math> (they are both supposed to be the midpoint of <math>MN</math>). In other words, since <math>\angle TP'N = 90^\circ</math>, then <math> \angle TPN = 90^\circ</math>. | ||
+ | |||
+ | Thus, <math>\angle TPR</math> subtends a <math>90^\circ \times 2 = 180^\circ</math> degree arc. So arc <math>TR</math> in circle <math>B</math> is <math>180^\circ</math>, so <math>TR</math> is a diameter, as desired. Thus <math>A</math>, <math>B</math>, <math>R</math> are collinear. | ||
== See Also == | == See Also == |
Revision as of 14:47, 24 February 2020
Problem
In the adjoining figure, two circles with radii and
are drawn with their centers
units apart. At
, one of the points of intersection, a line is drawn in such a way that the chords
and
have equal length. Find the square of the length of
.
Contents
[hide]Solution
Note that some of these solutions assume that lies on the line connecting the centers, which is not true in general. It is true here only because the perpendicular from
passes through through the point where the line between the centers intersects the small circle. This fact can be derived from the application of the Midpoint Theorem to the trapezoid made by dropping perpendiculars from the centers onto
.
Solution 1
Firstly, notice that if we reflect over
, we get
. Since we know that
is on circle
and
is on circle
, we can reflect circle
over
to get another circle (centered at a new point
, and with radius
) that intersects circle
at
. The rest is just finding lengths, as follows.
Since is the midpoint of segment
,
is a median of
. Because we know
,
, and
, we can find the third side of the triangle using Stewart's Theorem or similar approaches. We get
. Now we have a kite
with
,
, and
, and all we need is the length of the other diagonal
. The easiest way it can be found is with the Pythagorean Theorem. Let
be the length of
. Then

Solving this equation, we find that , so
Solution 2 (easiest)
Draw additional lines as indicated. Note that since triangles and
are isosceles, the altitudes are also bisectors, so let
.
Since triangles
and
are similar. If we let
, we have
.
Applying the Pythagorean Theorem on triangle , we have
. Similarly, for triangle
, we have
.
Subtracting, .
Solution 3
Let . Angles
,
, and
must add up to
. By the Law of Cosines,
. Also, angles
and
equal
and
. So we have

Taking the cosine of both sides, and simplifying using the addition formula for as well as the identity
, gives
.
Solution 4 (quickest)
Let . Extend the line containing the centers of the two circles to meet
, and to meet the other side of the large circle at a point
.
The part of this line from to the point nearest to
where it intersects the larger circle has length
.
The length of the diameter of the larger circle is
.
Thus by Power of a Point in the circle passing through ,
, and
, we have
, so
.
Full Proof that R, A, B are collinear
Let and
be the feet of the perpendicular from
to
and
to
respectively. It is well known that a perpendicular from the center of a circle to a chord of that circle bisects the chord, so
, since the problem told us
.
We will show that lies on
.
Let be the intersection of circle centered at
with
. Then
.
Let ' be the foot of the perpendicular from
to
. Then
is a midline (or midsegment) in trapezoid
, so
coincides with
(they are both supposed to be the midpoint of
). In other words, since
, then
.
Thus, subtends a
degree arc. So arc
in circle
is
, so
is a diameter, as desired. Thus
,
,
are collinear.
See Also
1983 AIME (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 |