Difference between revisions of "2016 AIME I Problems/Problem 6"
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In <math>\triangle ABC</math> let <math>I</math> be the center of the inscribed circle, and let the bisector of <math>\angle ACB</math> intersect <math>AB</math> at <math>L</math>. The line through <math>C</math> and <math>L</math> intersects the circumscribed circle of <math>\triangle ABC</math> at the two points <math>C</math> and <math>D</math>. If <math>LI=2</math> and <math>LD=3</math>, then <math>IC=\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | In <math>\triangle ABC</math> let <math>I</math> be the center of the inscribed circle, and let the bisector of <math>\angle ACB</math> intersect <math>AB</math> at <math>L</math>. The line through <math>C</math> and <math>L</math> intersects the circumscribed circle of <math>\triangle ABC</math> at the two points <math>C</math> and <math>D</math>. If <math>LI=2</math> and <math>LD=3</math>, then <math>IC=\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | ||
− | |||
==Solution 1== | ==Solution 1== | ||
Suppose we label the angles as shown below. | Suppose we label the angles as shown below. | ||
Line 47: | Line 46: | ||
==Solution 4== | ==Solution 4== | ||
− | Since <math>\angle{LAD} = \angle{BDC}</math> and <math>\angle{DLA}=\angle{ | + | Since <math>\angle{LAD} = \angle{BDC}</math> and <math>\angle{DLA}=\angle{DCB}</math>, <math>\triangle{DLA}\sim\triangle{DBC}</math>. Also, <math>\angle{DAC}=\angle{BLC}</math> and <math>\angle{ACD}=\angle{LCB}</math> so <math>\triangle{DAC}\sim\triangle{BLC}</math>. Now we can call <math>AC</math>, <math>b</math> and <math>BC</math>, <math>a</math>. By angle bisector theorem, <math>\frac{AD}{DB}=\frac{AC}{BC}</math>. So let <math>AD=bk</math> and <math>DB=ak</math> for some value of <math>k</math>. Now call <math>IC=x</math>. By the similar triangles we found earlier, <math>\frac{3}{ak}=\frac{bk}{x+2}</math> and <math>\frac{b}{x+5}=\frac{x+2}{a}</math>. We can simplify this to <math>abk^2=3x+6</math> and <math>ab=(x+5)(x+2)</math>. So we can plug the <math>ab</math> into the first equation and get <math>(x+5)(x+2)k^2=3(x+2) \rightarrow k^2(x+5)=3</math>. We can now draw a line through <math>A</math> and <math>I</math> that intersects <math>BC</math> at <math>E</math>. By mass points, we can assign a mass of <math>a</math> to <math>A</math>, <math>b</math> to <math>B</math>, and <math>a+b</math> to <math>D</math>. We can also assign a mass of <math>(a+b)k</math> to <math>C</math> by angle bisector theorem. So the ratio of <math>\frac{DI}{IC}=\frac{(a+b)k}{a+b}=k=\frac{2}{x}</math>. So since <math>k=\frac{2}{x}</math>, we can plug this back into the original equation to get <math>\left(\frac{2}{x}\right)^2(x+5)=3</math>. This means that <math>\frac{3x^2}{4}-x-5=0</math> which has roots -2 and <math>\frac{10}{3}</math> which means our <math>CI=\frac{10}{3}</math> and our answer is <math>\boxed{013}</math>. |
==Solution 5== | ==Solution 5== | ||
Line 56: | Line 55: | ||
~Pluto1708 | ~Pluto1708 | ||
− | Alternate solution: | + | Alternate solution: We can use the angle bisector theorem on <math>\triangle CBL</math> and bisector <math>BI</math> to get that <math>\tfrac{CI}{IL}=\tfrac{CI}{2}=\tfrac{BC}{BL}</math>. Since <math>\triangle CBL \sim \triangle ADL</math>, we get <math>\tfrac{BC}{BL}=\tfrac{AD}{DL}=\tfrac{5}{3}</math>. Thus, <math>CI=\tfrac{10}{3}</math> and <math>p+q=\boxed{13}</math>. |
(https://artofproblemsolving.com/community/c759169h1918283_geometry_problem) | (https://artofproblemsolving.com/community/c759169h1918283_geometry_problem) | ||
Line 64: | Line 63: | ||
==Solution 8 (Cyclic Quadrilaterals)== | ==Solution 8 (Cyclic Quadrilaterals)== | ||
− | + | <asy> | |
+ | size(150); | ||
+ | import olympiad; | ||
+ | real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6; | ||
+ | pair A=(0,0),B=(c,0),D=(c/2,-sqrt(25-(c/2)^2)); | ||
+ | pair C=intersectionpoints(circle(A,b),circle(B,a))[0]; | ||
+ | pair I=incenter(A,B,C); | ||
+ | pair L=extension(C,D,A,B); | ||
+ | dot(I^^A^^B^^C^^D); | ||
+ | draw(C--D); | ||
+ | path midangle(pair d,pair e,pair f) {return e--e+((f-e)/length(f-e)+(d-e)/length(d-e))/2;} | ||
+ | draw(A--B--D--cycle); | ||
+ | draw(circumcircle(A,B,D)); | ||
+ | draw(A--C--B); | ||
+ | draw(A--I--B^^C--I); | ||
+ | draw(incircle(A,B,C)); | ||
+ | label("$A$",A,SW,fontsize(8)); | ||
+ | label("$B$",B,SE,fontsize(8)); | ||
+ | label("$C$",C,N,fontsize(8)); | ||
+ | label("$D$",D,S,fontsize(8)); | ||
+ | label("$I$",I,NE,fontsize(8)); | ||
+ | label("$L$",L,SW,fontsize(8)); | ||
+ | label("$\alpha$",A,5*dir(midangle(C,A,I)),fontsize(8)); | ||
+ | label("$\alpha$",A,5*dir(midangle(I,A,B)),fontsize(8)); | ||
+ | label("$\beta$",B,12*dir(midangle(A,B,I)),fontsize(8)); | ||
+ | label("$\beta$",B,12*dir(midangle(I,B,C)),fontsize(8)); | ||
+ | label("$\gamma$",C,5*dir(midangle(A,C,I)),fontsize(8)); | ||
+ | label("$\gamma$",C,5*dir(midangle(I,C,B)),fontsize(8)); | ||
+ | </asy> | ||
+ | Connect <math>D</math> to <math>A</math> and <math>D</math> to <math>B</math> to form quadrilateral <math>ACBD</math>. Since quadrilateral <math>ACBD</math> is cyclic, we can apply Ptolemy's Theorem on the quadrilateral. | ||
+ | |||
+ | Denote the length of <math>BD</math> and <math>AD</math> as <math>z</math> (they must be congruent, as <math>\angle ABD</math> and <math>\angle DAB</math> are both inscribed in arcs that have the same degree measure due to the angle bisector intersecting the circumcircle at <math>D</math>), and the lengths of <math>BC</math>, <math>AC</math>, <math>AB</math>, and <math>CI</math> as <math>a,b,c, x</math>, respectively. | ||
+ | |||
+ | After applying Ptolemy's, one will get that: | ||
+ | |||
+ | <cmath>z(a+b)=c(x+5)</cmath> | ||
+ | |||
+ | Next, since <math>ACBD</math> is cyclic, triangles <math>ALD</math> and <math>CLB</math> are similar, yielding the following equation once simplifications are made to the equation <math>\frac{AD}{CB}=\frac{AL}{BL}</math>, with the length of <math>BL</math> written in terms of <math>a,b,c</math> using the angle bisector theorem on triangle <math>ABC</math>: | ||
+ | |||
+ | <cmath>zc=3(a+b)</cmath> | ||
+ | |||
+ | Next, drawing in the bisector of <math>\angle BAC</math> to the incenter <math>I</math>, and applying the angle bisector theorem, we have that: | ||
+ | |||
+ | <cmath>cx=2(a+b)</cmath> | ||
+ | |||
+ | Now, solving for <math>z</math> in the second equation, and <math>x</math> in the third equation and plugging them both back into the first equation, and making the substitution <math>w=\frac{a+b}{c}</math>, we get the quadratic equation: | ||
+ | |||
+ | <cmath>3w^2-2w-5=0</cmath> | ||
+ | |||
+ | Solving, we get <math>w=5/3</math>, which gives <math>z=5</math> and <math>x=10/3</math>, when we rewrite the above equations in terms of <math>w</math>. Thus, our answer is <math>\boxed{013}</math> and we're done. | ||
+ | |||
+ | -mathislife52 | ||
+ | |||
+ | ==Solution 9(Visual)== | ||
+ | [[File:2016 AIME I 6b.png|500px]] | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Solution 10== | ||
+ | Let <math>AB=c,BC=a,CA=b</math>, and <math>x=\tfrac{a+b}{c}</math>. Then, notice that <math>\tfrac{CI}{IL}=\tfrac{a+b}{c}=x</math>, so <math>CI=IL\cdot{}x=2x</math>. Also, by the incenter-excenter lemma, <math>AD=BD=ID=IL+LD=5</math>. Therefore, by Ptolemy's Theorem on cyclic quadrilateral <math>ABCD</math>, <math>5a+5b=c(2x+5)</math>, so <math>5\left(\tfrac{a+b}{c}\right)=2x+5</math>, so <math>5x=2x+5</math>. Solving, we get that <math>x=\tfrac{5}{3}</math>, so <math>CI=\tfrac{10}{3}</math> and the answer is <math>10+3=\boxed{013}</math>. | ||
+ | |||
+ | ==Solution 11== | ||
+ | |||
+ | Perform a <math>\sqrt{bc}</math> Inversion followed by a reflection along the angle bisector of <math>\angle BCA</math>. | ||
+ | |||
+ | It's well known that | ||
+ | <cmath>AB \leftrightarrow \odot CBA \implies L \leftrightarrow D</cmath> | ||
+ | <cmath>I \leftrightarrow I_A</cmath> | ||
+ | where <math>I_A</math> is the <math>A-</math>excenter. | ||
+ | |||
+ | Also by Fact 5, <math>DI_A = 5</math>. | ||
+ | |||
+ | So, | ||
+ | <cmath>CL \cdot CD = CI \cdot CI_A</cmath> | ||
+ | <cmath>\implies (CI + IL) \cdot (CI + ID) = (CI) \cdot (CI + II_A)</cmath> | ||
+ | <cmath>\implies (CI + 2) \cdot (CI + 5) = (CI) \cdot (CI + 10)</cmath> | ||
+ | <cmath>\implies 7CI +10= 10CI</cmath> | ||
+ | <cmath>\implies CI = \boxed{\dfrac{10}{3}}.\blacksquare</cmath> | ||
+ | |||
+ | ~kamatadu | ||
+ | |||
+ | |||
+ | ==Solution 12== | ||
+ | Without loss of generality, let <math>\triangle ABC</math> be isosceles. Note that by the incenter-excenter lemma, <math>DI = DA = DB.</math> Hence, <math>DA=DB=5.</math> Let the point of tangency of the incircle and <math>\overline{BC}</math> be <math>F</math> and the point of tangency of the incircle and <math>\overline{AC}</math> be <math>E.</math> We note that <math>\angle ALC = \angle BLC = 90^\circ</math> and <math>LA=LB=4,</math> which immediately gives <math>AE=BF=4.</math> Applying the Pythagorean Theorem on <math>\triangle ALC</math> and <math>\triangle IEC</math> gives <math>2^2+x^2=y^2</math> and <math>4^2+(2+y)^2 = (4+x)^2.</math> Solving for <math>y</math> gives us <math>y=\frac{10}{3}.</math> Therefore, <math>IC = \frac{10}{3}</math> so the answer is <math>\boxed{13}.</math> | ||
+ | |||
+ | ~peelybonehead | ||
+ | |||
+ | ==Solution 13 (Trig)== | ||
+ | Let <math>C_1\in AB</math> be the point such that <math>IC_1\perp AB</math>, and let <math>D_1\in AB</math> be defined similarly for <math>D</math>. We know that <math>\triangle IC_1L\sim\triangle DD_1L</math>, so by triangle ratios <math>DD_1=\frac{3}{2}r</math>, where <math>r</math> is the inradius. Additionally, by cyclic quadrilaterals, we know that <math>\angle BAD=\angle DAB=\frac{\gamma}{2}</math>, where <math>\gamma</math> is equivalent to <math>\angle ACB</math>. Thus <math>\triangle ADB</math> is isosceles and <math>DD_1</math> is the perpendicular bisector of the triangle, so <math>AD_1=\frac{c}{2}</math>. Since <math>\tan\left(\frac{\gamma}{2}\right)=\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}</math> from formulas (where <math>s</math> is half the perimeter of <math>\triangle ABC</math>) and since <math>\tan\left(\frac{\gamma}{2}\right)=\frac{\frac{3}{2}r}{\frac{1}{2}c}</math> from <math>\triangle ADB</math>, we can set up an equation: | ||
+ | |||
+ | <math>\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}=\frac{\frac{3}{2}r}{\frac{1}{2}c}\implies\frac{\sqrt{s(s-a)(s-b)(s-c)}}{s(s-c)}=\frac{3rs}{cs}\implies\frac{[ABC]}{s(s-c)}=\frac{3[ABC]}{cs}</math> | ||
+ | |||
+ | <math>\implies\frac{1}{s-c}=\frac{3}{c}\implies \frac{s}{c}=\frac{4}{3}</math> | ||
+ | |||
+ | Let <math>C_2\in AB</math> such that <math>CC_2\perp AB</math>. Then <math>CC_2=\frac{2[ABC]}{c}</math>. Using the area formula <math>[ABC]=rs</math> and our fact from above yields <math>CC_2=\frac{8}{3}r</math>. We then notice that <math>\triangle CC_2L\sim\triangle IC_1L</math>, so if we let <math>x=CI</math>, by triangle ratios we find that <math>\frac{\frac{8}{3}r}{x+2}=\frac{r}{2}</math>, leading to <math>x=\frac{10}{3}</math>. Thus the answer is <math>10+3=\boxed{013}</math>. | ||
+ | |||
+ | ~eevee9406 | ||
== See also == | == See also == | ||
{{AIME box|year=2016|n=I|num-b=5|num-a=7}} | {{AIME box|year=2016|n=I|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 13:36, 16 July 2024
Contents
Problem
In let
be the center of the inscribed circle, and let the bisector of
intersect
at
. The line through
and
intersects the circumscribed circle of
at the two points
and
. If
and
, then
, where
and
are relatively prime positive integers. Find
.
Solution 1
Suppose we label the angles as shown below.
As
and
intercept the same arc, we know that
. Similarly,
. Also, using
, we find
. Therefore,
. Therefore,
, so
must be isosceles with
. Similarly,
. Then
, hence
. Also,
bisects
, so by the Angle Bisector Theorem
. Thus
, and the answer is
.
Solution 2
WLOG assume is isosceles. Then,
is the midpoint of
, and
. Draw the perpendicular from
to
, and let it meet
at
. Since
,
is also
(they are both inradii). Set
as
. Then, triangles
and
are similar, and
. Thus,
.
, so
. Thus
. Solving for
, we have:
, or
.
is positive, so
. As a result,
and the answer is
Solution 3
WLOG assume is isosceles (with vertex
). Let
be the center of the circumcircle,
the circumradius, and
the inradius. A simple sketch will reveal that
must be obtuse (as an acute triangle will result in
being greater than
) and that
and
are collinear. Next, if
,
and
. Euler gives us that
, and in this case,
. Thus,
. Solving for
, we have
, then
, yielding
. Next,
so
. Finally,
gives us
, and
. Our answer is then
.
Solution 4
Since and
,
. Also,
and
so
. Now we can call
,
and
,
. By angle bisector theorem,
. So let
and
for some value of
. Now call
. By the similar triangles we found earlier,
and
. We can simplify this to
and
. So we can plug the
into the first equation and get
. We can now draw a line through
and
that intersects
at
. By mass points, we can assign a mass of
to
,
to
, and
to
. We can also assign a mass of
to
by angle bisector theorem. So the ratio of
. So since
, we can plug this back into the original equation to get
. This means that
which has roots -2 and
which means our
and our answer is
.
Solution 5
Since and
both intercept arc
, it follows that
. Note that
by the external angle theorem. It follows that
, so we must have that
is isosceles, yielding
. Note that
, so
. This yields
. It follows that
, giving a final answer of
.
Solution 6
Let be the excenter opposite to
in
. By the incenter-excenter lemma
. Its well known that
.
~Pluto1708
Alternate solution: We can use the angle bisector theorem on and bisector
to get that
. Since
, we get
. Thus,
and
.
(https://artofproblemsolving.com/community/c759169h1918283_geometry_problem)
Solution 7
We can just say that quadrilateral is a right kite with right angles at
and
. Let us construct another similar right kite with the points of tangency on
and
called
and
respectively, point
, and point
. Note that we only have to look at one half of the circle since the diagram is symmetrical. Let us call
for simplicity's sake. Based on the fact that
is similar to
we can use triangle proportionality to say that
is
. Using geometric mean theorem we can show that
must be
. With Pythagorean Theorem we can say that
. Multiplying both sides by
and moving everything to LHS will give you
Since
must be in the form
we can assume that
is most likely a positive fraction in the form
where
is a factor of
. Testing the factors in synthetic division would lead
, giving us our desired answer
. ~Lopkiloinm
Solution 8 (Cyclic Quadrilaterals)
Connect
to
and
to
to form quadrilateral
. Since quadrilateral
is cyclic, we can apply Ptolemy's Theorem on the quadrilateral.
Denote the length of and
as
(they must be congruent, as
and
are both inscribed in arcs that have the same degree measure due to the angle bisector intersecting the circumcircle at
), and the lengths of
,
,
, and
as
, respectively.
After applying Ptolemy's, one will get that:
Next, since is cyclic, triangles
and
are similar, yielding the following equation once simplifications are made to the equation
, with the length of
written in terms of
using the angle bisector theorem on triangle
:
Next, drawing in the bisector of to the incenter
, and applying the angle bisector theorem, we have that:
Now, solving for in the second equation, and
in the third equation and plugging them both back into the first equation, and making the substitution
, we get the quadratic equation:
Solving, we get , which gives
and
, when we rewrite the above equations in terms of
. Thus, our answer is
and we're done.
-mathislife52
Solution 9(Visual)
vladimir.shelomovskii@gmail.com, vvsss
Solution 10
Let , and
. Then, notice that
, so
. Also, by the incenter-excenter lemma,
. Therefore, by Ptolemy's Theorem on cyclic quadrilateral
,
, so
, so
. Solving, we get that
, so
and the answer is
.
Solution 11
Perform a Inversion followed by a reflection along the angle bisector of
.
It's well known that
where
is the
excenter.
Also by Fact 5, .
So,
~kamatadu
Solution 12
Without loss of generality, let be isosceles. Note that by the incenter-excenter lemma,
Hence,
Let the point of tangency of the incircle and
be
and the point of tangency of the incircle and
be
We note that
and
which immediately gives
Applying the Pythagorean Theorem on
and
gives
and
Solving for
gives us
Therefore,
so the answer is
~peelybonehead
Solution 13 (Trig)
Let be the point such that
, and let
be defined similarly for
. We know that
, so by triangle ratios
, where
is the inradius. Additionally, by cyclic quadrilaterals, we know that
, where
is equivalent to
. Thus
is isosceles and
is the perpendicular bisector of the triangle, so
. Since
from formulas (where
is half the perimeter of
) and since
from
, we can set up an equation:
Let such that
. Then
. Using the area formula
and our fact from above yields
. We then notice that
, so if we let
, by triangle ratios we find that
, leading to
. Thus the answer is
.
~eevee9406
See also
2016 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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