Difference between revisions of "2019 AIME I Problems/Problem 15"
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==Solution 4== | ==Solution 4== | ||
− | Note that the tangents to the circles at <math>A</math> and <math>B</math> intersect at a point <math>Z</math> on <math>XY</math> by radical | + | [[File:AIME-I-2019-15.png|250px|right]] |
+ | Note that the tangents to the circles at <math>A</math> and <math>B</math> intersect at a point <math>Z</math> on <math>XY</math> by radical axis theorem. Since <math>\angle ZAB = \angle ZQA</math> and <math>\angle ZBA = \angle ZQB</math>, we have | ||
<cmath>\angle AZB + \angle AQB = \angle AZB + \angle ZAB + \angle ZBA = 180^{\circ},</cmath> | <cmath>\angle AZB + \angle AQB = \angle AZB + \angle ZAB + \angle ZBA = 180^{\circ},</cmath> | ||
− | so <math>ZAQB</math> is cyclic. But if <math>O</math> is the center of <math>\omega</math>, clearly <math>ZAOB</math> is cyclic with diameter <math>ZO</math>, so <math>\angle ZQO = 90^{\circ} | + | so <math>ZAQB</math> is cyclic. |
+ | |||
+ | But if <math>O</math> is the center of <math>\omega</math>, clearly <math>ZAOB</math> is cyclic with diameter <math>ZO</math>, so <math>\angle ZQO = 90^{\circ}</math> implies that <math>Q</math> is the midpoint of <math>XY</math>. Then, by power of point <math>P</math>, <cmath>PY \cdot PX = PA \cdot PB = 15,</cmath> whereas it is given that <math>PY+PX = 11</math>. Thus <cmath>PY, PX \in \left\{\tfrac 12 (11 \pm \sqrt{61})\right\}</cmath> so <math>PQ = \frac{\sqrt{61}}{2}</math>, i.e. <math>PQ^2 = \frac{61}{4}</math> and the answer is <math>61+4 = \boxed{065}</math>. | ||
+ | |||
+ | ==Solution 5== | ||
+ | |||
+ | Connect <math>AQ,QB</math>, since <math>\angle{AO_1P}=\angle{AOB}=\angle{BO_2P}</math>, so <math>\angle{AQP}=\frac{\angle{AO_1P}}{2}=\angle{BQP}=\frac{\angle{BO_2P}}{2}, \angle{AQB}=\angle{AOB}</math> then, so <math>A,O,Q,B</math> are concyclic | ||
+ | |||
+ | We let <math>\angle{AO_1P}=\angle{AOB}=\angle{BO_2P}=2\alpha</math>, it is clear that <math>\angle{BQP}=\alpha, \angle{O_1AP}=90^{\circ}-\alpha</math>, which leads to the conclusion <math>OQ\bot XY</math> which tells <math>Q</math> is the midpoint of <math>XY</math> | ||
+ | |||
+ | Then it is clear, <math>XP\cdot PY=15, XP=\frac{11-\sqrt{61}}{2}, PQ=\frac{11}{2}-\frac{11-\sqrt{61}}{2}=\frac{\sqrt{61}}{2}</math> , the answer is <math>\boxed{065}</math> | ||
+ | |||
+ | ~bluesoul | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | ==Solution 6(lazy)== | ||
+ | <asy> | ||
+ | size(8cm); | ||
+ | pair O, A, B, P, O1, O2, Q, X, Y; | ||
+ | O=(0, 0); | ||
+ | A=dir(140); B=dir(40); | ||
+ | P=(3A+5B)/8; | ||
+ | O1=extension((A+P)/2, (A+P)/2+(0, 1), A, O); | ||
+ | O2=extension((B+P)/2, (B+P)/2+(0, 1), B, O); | ||
+ | Q=intersectionpoints(circle(O1, length(A-O1)), circle(O2, length(B-O2)))[1]; | ||
+ | X=intersectionpoint(P -- (P+(P-Q)*100), circle(O, 1)); | ||
+ | Y=intersectionpoint(Q -- (Q+(Q-P)*100), circle(O, 1)); | ||
+ | |||
+ | draw(circle(O, 1)); | ||
+ | draw(circle(O1, length(A-O1))); | ||
+ | draw(circle(O2, length(B-O2))); | ||
+ | draw(A -- B,red); draw(X -- Y,green); | ||
+ | |||
+ | dot("$A$", A, A); | ||
+ | dot("$B$", B, B); | ||
+ | dot("$P$", P, dir(70),blue); | ||
+ | dot("$Q$", Q, dir(200)); | ||
+ | dot("$X$", X, X); | ||
+ | dot("$Y$", Y, Y); | ||
+ | label("$3$", (A+P)/2, N, red); | ||
+ | label("$5$", (B+P)/2, N, red); | ||
+ | draw(brace(X,Y)); | ||
+ | label("$11$",brace(X,Y),dir(20)); | ||
+ | </asy> | ||
+ | <math>PX \cdot PY=AP \cdot PB=5 \cdot 3=15</math> by power of a point. Also, <math>PX+PY=XY=11</math>, so <math>PX</math> and <math>PY</math> are solutions to the quadratic <math>x^2-11x+15=0</math> so <math>PX</math> and <math>PY</math> is <math>\frac{11\pm\sqrt{61}}{2}</math> in some order. Now, because we want <math>PQ^2</math> and it is known to be rational, we can guess that <math>PQ</math> is irrational or the problem would simply ask for <math>PQ</math>. We can also figure out that since <math>PQ^2</math> is rational, <math>PQ</math> is <math>\sqrt{\text{[something]}}</math>. <math>PQ=QX-PX</math>, and chances are low that <math>QX</math> is some number with a square root plus or minus <math>\frac{\sqrt{61}}{2}</math> to cancel out the <math>\frac{\sqrt{61}}{2}</math> in <math>PX</math>, so one can see that <math>PQ^2</math> is most likely to be <math>\left(\frac{\sqrt{61}}{2}\right)^2=\frac{61}{4}</math>, and our answer is <math>61+4=\boxed{065}</math> | ||
+ | |||
+ | Note : If our answer is correct, then <math>QX=\frac{11}{2}</math>, which made <math>Q</math> the midpoint of <math>XY</math>, a feature that occurs often in AIME problems, so that again made our answer probable. Midpoints have many properties and there is a lot of ways to show if a point is the midpoint of a segment. Even if the answer is wrong, it's still the same as leaving it blank and 065 is a good guess. ~[[Ddk001]] | ||
+ | |||
+ | ==Solution 7== | ||
+ | We will show that <math>Q</math> is the midpoint of <math>XY.</math> To do this, let <math>Q^{\prime}</math> be the altitude from <math>O</math> to <math>XY</math> or, equivalently, to <math>PQ.</math> Notice that <math>O_{1}00_{2}P</math> is a parallelogram. Thus, the height from <math>O</math> to <math>O_{1}O_{2}</math> is equal to the height from <math>P</math> to <math>O_{1}O_{2}.</math> Say that the line through <math>P</math> perpendicular to <math>O_{1}O_{2}</math> intersects <math>O_{1}O_{2}</math> at <math>H.</math> Then, <math>PQ</math> is perpendicular to <math>O_{1}O_{2},</math> so <math>H</math> is on <math>PQ.</math> Now, we have that the altitude from <math>O</math> to <math>O_{1}O_{2}</math> is equal to the altitude from <math>Q^{\prime}</math> to <math>O_{1}O_{2}</math> (since <math>OQ^{\prime} \parallel O_{1}O_{2}</math>). However, the altitude from <math>Q^{\prime}</math> to <math>O_{1}O_{2}</math> is just <math>Q^{\prime}H.</math> Also, the altitude from <math>P</math> to <math>O_{1}O_{2}</math> is <math>PH</math>, so <math>PH = Q^{\prime}H.</math> Thus, <math>O_{1}O_{2}</math> bisects <math>PQ^{\prime}.</math> However, this is true for <math>Q,</math> too, so <math>Q = Q^{\prime},</math> and we are done. Now, by PoP, we have | ||
+ | <cmath>AP \cdot BP = XP \cdot YP = 15.</cmath> | ||
+ | Also, we have <math>XY = XP+YP = 11,</math> so <math>XP = \frac{11 \pm \sqrt{61}}{2}</math>. Notice that <math>XQ = \frac{XY}{2} = \frac{11}{2},</math> so <math>PQ = \frac{\sqrt{61}}{2},</math> giving us our answer of <math>\boxed{065}.</math> | ||
+ | |||
+ | |||
+ | ==Solution 8== | ||
+ | |||
+ | [[File:2019_AIME_Problem_15_Diagram.png|400px|thumb|right|[https://www.overleaf.com/read/bkdngbhwcskc#415c3f Latex]]] | ||
+ | |||
+ | Like Solution 7, let <math>Q'</math> be the altitude from <math>O</math> to <math>XY</math>. And, let <math>M</math> be the intersection of <math>O_1O_2</math> and <math>PQ</math>. Construct <math>P'</math> on line <math>AO</math> such that <math>PP' \parallel O_2O_1</math>. First, because of isosceles triangles <math>OAB</math>, <math>O_1AP</math>, and <math>O_2BP</math>, we have <math>\angle{OAP} = \angle{OBA} = \angle{APO_1} = \angle{BPO_2}</math>, which means <math>OO_1PO_2</math> is a parallelogram. So, <math>O_2P = OO_1</math>. It is also clear that <math>PP'O_1O_2</math> is a parallelogram by virtue of our definition. Thus, <math>O_2P = O_1P' = OO_1</math>. Since <math>OQ' \parallel O_1O_2 \parallel P'P</math> (because of the right angles), <math>\frac{Q'M}{MP} = \frac{OO_1}{O_1P'} = 1 \implies Q'M = MP</math>. And, because <math>QM = MP</math>, <math>Q = Q'</math>. From Power of a Point on <math>P</math>, we have <math>XP(11-XP) = 15</math>, giving us <math>XP = \frac{11 - \sqrt{61}}{2}</math>. Since <math>OQ</math> is perpendicular to <math>XY</math>, <math>Q</math> is the midpoint of <math>XY</math>, so <math>XQ = \frac{11}{2}</math>. Thus, <math>PQ = \frac{11}{2} - \frac{11 - \sqrt{61}}{2} = \frac{\sqrt{61}}{2} \implies {PQ}^2 = \frac{61}{4}</math>. Therefore, our answer is <math>\boxed{65}</math>. | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Crazyvideogamez CrazyVideoGamez] | ||
+ | |||
+ | <math> | ||
+ | \newline | ||
+ | </math> | ||
+ | |||
+ | ==Video Solution by Mr. Math== | ||
+ | |||
+ | https://www.youtube.com/watch?v=X_CSRwUh0Rc | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2019|n=I|num-b=14|after=Last Problem}} | {{AIME box|year=2019|n=I|num-b=14|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] |
Latest revision as of 11:32, 1 February 2024
Contents
Problem
Let be a chord of a circle
, and let
be a point on the chord
. Circle
passes through
and
and is internally tangent to
. Circle
passes through
and
and is internally tangent to
. Circles
and
intersect at points
and
. Line
intersects
at
and
. Assume that
,
,
, and
, where
and
are relatively prime positive integers. Find
.
Solution 1
Let
and
be the centers of
and
, respectively. There is a homothety at
sending
to
that sends
to
and
to
, so
. Similarly,
, so
is a parallelogram. Moreover,
whence
is cyclic. However,
so
is an isosceles trapezoid. Since
,
, so
is the midpoint of
.
By Power of a Point, . Since
and
,
and the requested sum is
.
(Solution by TheUltimate123)
Note
One may solve for first using PoAP,
. Then, notice that
is rational but
is not, also
. The most likely explanation for this is that
is the midpoint of
, so that
and
. Then our answer is
. One can rigorously prove this using the methods above
Solution 2
Let the tangents to at
and
intersect at
. Then, since
,
lies on the radical axis of
and
, which is
. It follows that
Let
denote the midpoint of
. By the Midpoint of Harmonic Bundles Lemma(EGMO 9.17),
whence
. Like above,
. Since
, we establish that
, from which
, and the requested sum is
.
(Solution by TheUltimate123)
Solution 3
Firstly we need to notice that is the middle point of
. Assume the center of circle
are
, respectively. Then
are collinear and
are collinear. Link
. Notice that,
. As a result,
and
. So we have parallelogram
. So
Notice that,
and
divides
into two equal length pieces, So we have
. As a result,
lie on one circle. So
. Notice that since
, we have
. As a result,
. So
is the middle point of
.
Back to our problem. Assume ,
and
. Then we have
, that is,
. Also,
. Solve these above, we have
. As a result, we have
. So, we have
. As a result, our answer is
.
Solution By BladeRunnerAUG (Fanyuchen20020715). Edited by bgn4493.
Solution 4
Note that the tangents to the circles at and
intersect at a point
on
by radical axis theorem. Since
and
, we have
so
is cyclic.
But if is the center of
, clearly
is cyclic with diameter
, so
implies that
is the midpoint of
. Then, by power of point
,
whereas it is given that
. Thus
so
, i.e.
and the answer is
.
Solution 5
Connect , since
, so
then, so
are concyclic
We let , it is clear that
, which leads to the conclusion
which tells
is the midpoint of
Then it is clear, , the answer is
~bluesoul
Solution 6(lazy)
by power of a point. Also,
, so
and
are solutions to the quadratic
so
and
is
in some order. Now, because we want
and it is known to be rational, we can guess that
is irrational or the problem would simply ask for
. We can also figure out that since
is rational,
is
.
, and chances are low that
is some number with a square root plus or minus
to cancel out the
in
, so one can see that
is most likely to be
, and our answer is
Note : If our answer is correct, then , which made
the midpoint of
, a feature that occurs often in AIME problems, so that again made our answer probable. Midpoints have many properties and there is a lot of ways to show if a point is the midpoint of a segment. Even if the answer is wrong, it's still the same as leaving it blank and 065 is a good guess. ~Ddk001
Solution 7
We will show that is the midpoint of
To do this, let
be the altitude from
to
or, equivalently, to
Notice that
is a parallelogram. Thus, the height from
to
is equal to the height from
to
Say that the line through
perpendicular to
intersects
at
Then,
is perpendicular to
so
is on
Now, we have that the altitude from
to
is equal to the altitude from
to
(since
). However, the altitude from
to
is just
Also, the altitude from
to
is
, so
Thus,
bisects
However, this is true for
too, so
and we are done. Now, by PoP, we have
Also, we have
so
. Notice that
so
giving us our answer of
Solution 8
Like Solution 7, let be the altitude from
to
. And, let
be the intersection of
and
. Construct
on line
such that
. First, because of isosceles triangles
,
, and
, we have
, which means
is a parallelogram. So,
. It is also clear that
is a parallelogram by virtue of our definition. Thus,
. Since
(because of the right angles),
. And, because
,
. From Power of a Point on
, we have
, giving us
. Since
is perpendicular to
,
is the midpoint of
, so
. Thus,
. Therefore, our answer is
.
Video Solution by Mr. Math
https://www.youtube.com/watch?v=X_CSRwUh0Rc
See Also
2019 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Problem | |
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.