Difference between revisions of "2001 USAMO Problems/Problem 2"
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== Problem == | == Problem == | ||
Let <math>ABC</math> be a [[triangle]] and let <math>\omega</math> be its [[incircle]]. Denote by <math>D_1</math> and <math>E_1</math> the points where <math>\omega</math> is tangent to sides <math>BC</math> and <math>AC</math>, respectively. Denote by <math>D_2</math> and <math>E_2</math> the points on sides <math>BC</math> and <math>AC</math>, respectively, such that <math>CD_2 = BD_1</math> and <math>CE_2 = AE_1</math>, and denote by <math>P</math> the point of intersection of segments <math>AD_2</math> and <math>BE_2</math>. Circle <math>\omega</math> intersects segment <math>AD_2</math> at two points, the closer of which to the vertex <math>A</math> is denoted by <math>Q</math>. Prove that <math>AQ = D_2P</math>. | Let <math>ABC</math> be a [[triangle]] and let <math>\omega</math> be its [[incircle]]. Denote by <math>D_1</math> and <math>E_1</math> the points where <math>\omega</math> is tangent to sides <math>BC</math> and <math>AC</math>, respectively. Denote by <math>D_2</math> and <math>E_2</math> the points on sides <math>BC</math> and <math>AC</math>, respectively, such that <math>CD_2 = BD_1</math> and <math>CE_2 = AE_1</math>, and denote by <math>P</math> the point of intersection of segments <math>AD_2</math> and <math>BE_2</math>. Circle <math>\omega</math> intersects segment <math>AD_2</math> at two points, the closer of which to the vertex <math>A</math> is denoted by <math>Q</math>. Prove that <math>AQ = D_2P</math>. | ||
+ | |||
+ | ==Solution== | ||
+ | ===Solution 1=== | ||
+ | |||
+ | It is well known that the excircle opposite <math>A</math> is tangent to <math>\overline{BC}</math> at the point <math>D_2</math>. (Proof: let the points of tangency of the excircle with the lines <math>BC, AB, AC</math> be <math>D_3, F,G</math> respectively. Then <math>AB+BD_3=AB + BF=AF = AG = AC + CG=AC + CD_3</math>. It follows that <math>2CD_3 = AB + BC - AC</math>, and <math>CD_3 = s-b = BD_1 = CD_2</math>, so <math>D_3 \equiv D_2</math>.) | ||
+ | |||
+ | Now consider the homothety that carries the incircle of <math>\triangle ABC</math> to its excircle. The homothety also carries <math>Q</math> to <math>D_2</math> (since <math>A,Q,D_2</math> are collinear), and carries the tangency points <math>E_1</math> to <math>G</math>. It follows that <math>\frac{AQ}{QD_2} = \frac{AE_1}{E_1G} = \frac{s-a}{E_1C + CD_2} = \frac{s-a}{CD_1 + BD_1} = \frac{s-a}{a}</math>. | ||
+ | |||
+ | <asy> | ||
+ | pathpen = linewidth(0.7); size(300); pen d = linetype("4 4") + linewidth(0.6); pair B=(0,0), C=(10,0), A=7*expi(1),O=D(incenter(A,B,C)),D1 = D(MP("D_1",foot(O,B,C))),E1 = D(MP("E_1",foot(O,A,C),NE)),E2 = D(MP("E_2",C+A-E1,NE)); /* arbitrary points */ /* ugly construction for OA */ pair Ca = 2C-A, Cb = bisectorpoint(Ca,C,B), OA = IP(A--A+10*(O-A),C--C+50*(Cb-C)), D2 = D(MP("D_2",foot(OA,B,C))), Fa=2B-A, Ga=2C-A, F=MP("F",D(foot(OA,B,Fa)),NW), G=MP("G",D(foot(OA,C,Ga)),NE); D(OA); D(MP("A",A,N)--MP("B",B,NW)--MP("C",C,NE)--cycle); D(incircle(A,B,C)); D(CP(OA,D2),d); D(B--Fa,linewidth(0.6)); D(C--Ga,linewidth(0.6)); D(MP("P",IP(D(A--D2),D(B--E2)),NNE)); D(MP("Q",IP(incircle(A,B,C),A--D2),SW)); clip((-20,-10)--(-20,20)--(20,20)--(20,-10)--cycle); </asy> | ||
+ | By Menelaus' Theorem on <math>\triangle ACD_2</math> with segment <math>\overline{BE_2}</math>, it follows that <math>\frac{CE_2}{E_2A} \cdot \frac{AP}{PD_2} \cdot \frac{BD_2}{BC} = 1 \Longrightarrow \frac{AP}{PD_2} = \frac{(c - (s-a)) \cdot a}{(a-(s-c)) \cdot AE_1} = \frac{a}{s-a}</math>. It easily follows that <math>AQ = D_2P</math>. <math>\blacksquare</math> | ||
+ | |||
+ | ===Solution 2=== | ||
+ | |||
+ | The key observation is the following lemma. | ||
+ | |||
+ | Lemma: Segment <math>D_1Q</math> is a diameter of circle <math>\omega</math>. | ||
+ | |||
+ | Proof: Let <math>I</math> be the center of circle <math>\omega</math>, i.e., <math>I</math> is the incenter of triangle <math>ABC</math>. Extend segment <math>D_1I</math> through <math>I</math> to intersect circle <math>\omega</math> again at <math>Q'</math>, and extend segment <math>AQ'</math> through <math>Q'</math> to intersect segment <math>BC</math> at <math>D'</math>. We show that <math>D_2 = D'</math>, which in turn implies that <math>Q = Q'</math>, that is, <math>D_1Q</math> is a diameter of <math>\omega</math>. | ||
+ | |||
+ | Let <math>l</math> be the line tangent to circle <math>\omega</math> at <math>Q'</math>, and let <math>l</math> intersect the segments <math>AB</math> and <math>AC</math> at <math>B_1</math> and <math>C_1</math>, respectively. Then <math>\omega</math> is an excircle of triangle <math>AB_1C_1</math>. Let <math>\mathbf{H}_1</math> denote the dilation with its center at <math>A</math> and ratio <math>AD'/AQ'</math>. Since <math>l\perp D_1Q'</math> and <math>BC\perp D_1Q'</math>, <math>l\parallel BC</math>. Hence <math>AB/AB_1 = AC/AC_1 = AD'/AQ'</math>. Thus <math>\mathbf{H}_1(Q') = D'</math>, <math>\mathbf{H}_1(B_1) = B</math>, and <math>\mathbf{H}_1(C_1) = C</math>. It also follows that an excircle <math>\Omega</math> of triangle <math>ABC</math> is tangent to the side <math>BC</math> at <math>D'</math>. | ||
+ | |||
+ | It is well known that <cmath>CD_1 = \frac{1}{2}(BC + CA - AB).</cmath> We compute <math>BD'</math>. Let <math>X</math> and <math>Y</math> denote the points of tangency of circle <math>\Omega</math> with rays <math>AB</math> and <math>AC</math>, respectively. Then by equal tangents, <math>AX = AY</math>, <math>BD' = BX</math>, and <math>D'C = YC</math>. Hence <cmath>AX = AY = \frac{1}{2}(AX + AY) = \frac{1}{2}(AB + BX + YC + CA) = \frac{1}{2}(AB + BC + CA).</cmath> It follows that <cmath>BD' = BX = AX - AB = \frac{1}{2}(BC + CA - AB).</cmath> Combining these two equations yields <math>BD' = CD_1</math>. Thus <cmath>BD_2 = BD_1 - D_2D_1 = D_2C - D_2D_1 = CD_1 = BD',</cmath> that is, <math>D' = D_2</math>, as desired. <math>\blacksquare</math> | ||
+ | |||
+ | Now we prove our main result. Let <math>M_1</math> and <math>M_2</math> be the respective midpoints of segments <math>BC</math> and <math>CA</math>. Then <math>M_1</math> is also the midpoint of segment <math>D_1D_2</math>, from which it follows that <math>IM_1</math> is the midline of triangle <math>D_1QD_2</math>. Hence <cmath>QD_2 = 2IM_1</cmath> and <math>AD_2\parallel M_1I</math>. Similarly, we can prove that <math>BE_2\parallel M_2I</math>. | ||
+ | |||
+ | 2001usamo2-2.png | ||
+ | Let <math>G</math> be the centroid of triangle <math>ABC</math>. Thus segments <math>AM_1</math> and <math>BM_2</math> intersect at <math>G</math>. Define transformation <math>\mathbf{H}_2</math> as the dilation with its center at <math>G</math> and ratio <math>-1/2</math>. Then <math>\mathbf{H}_2(A) = M_1</math> and <math>\mathbf{H}_2(B) = M_2</math>. Under the dilation, parallel lines go to parallel lines and the intersection of two lines goes to the intersection of their images. Since <math>AD_2\parallel M_1I</math> and <math>BE_2\parallel M_2I</math>, <math>\mathbf{H}_2</math> maps lines <math>AD_2</math> and <math>BE_2</math> to lines <math>M_1I</math> and <math>M_2I</math>, respectively. It also follows that <math>\mathbf{H}_2(I) = P</math> and <cmath>\frac{IM_1}{AP} = \frac{GM_1}{AG} = \frac{1}{2}</cmath> or <cmath>AP = 2IM_1.</cmath> This yields <cmath>AQ = AP - QP = 2IM_1 - QP = QD_2 - QP = PD_2,</cmath> as desired. | ||
+ | |||
+ | Note: We used directed lengths in our calculations to avoid possible complications caused by the different shapes of triangle <math>ABC</math>. | ||
+ | |||
+ | === Solution 3 === | ||
+ | |||
+ | Here is a rather nice solution using barycentric coordinates: | ||
+ | |||
+ | Let <math>A</math> be <math>(1,0,0)</math>, <math>B</math> be <math>(0,1,0)</math>, and <math>C</math> be <math>(0,0,1)</math>. Let the side lengths of the triangle be <math>a,b,c</math> and the semi-perimeter <math>s</math>. | ||
+ | |||
+ | Now, <cmath>CD_1=s-c, BD_1=s-b, AE_1=s-a, CE_1=s-c.</cmath> Thus, <cmath>CD_2=s-b, BD_2=s-c, AE_2=s-c, CE_2=s-a.</cmath> | ||
+ | |||
+ | Therefore, <math>D_2=(0:s-b:s-c)</math> and <math>E_2=(s-a:0:s-c).</math> Clearly then, the non-normalized coordinates of <math>P=(s-a:s-b:s-c).</math> | ||
+ | |||
+ | Normalizing, we have that <cmath>D_2=\left(0,\frac{s-b}{a},\frac{s-c}{a}\right), E_2=\left(\frac{s-a}{b},0,\frac{s-c}{b}\right), P=\left(\frac{s-a}{s},\frac{s-b}{s},\frac{s-c}{s}\right).</cmath> | ||
+ | |||
+ | Now, we find the point <math>Q'</math> inside the triangle on the line <math>AD_2</math> such that <math>AQ'=D_2P</math>. It is then sufficient to show that this point lies on the incircle. | ||
+ | |||
+ | <math>P</math> is the fraction <math>\frac{s-a}{s}</math> of the way "up" the line segment from <math>D_2</math> to <math>A</math>. Thus, we are looking for the point that is <math>\frac{s-a}{s}</math> of the way "down" the line segment from <math>A</math> to <math>D_2</math>, or, the fraction <math>1-\frac{s-a}{s}</math> of the way "up". | ||
+ | |||
+ | Thus, <math>Q'</math> has normalized <math>x</math>-coordinate <math>1-\frac{s-a}{s}=\frac{a}{s}</math>. | ||
+ | |||
+ | As the line <math>AD_2</math> has equation <math>(s-c)y=(s-b)z</math>, it can easily be found that <cmath>Q'=\left(\frac{a^2}{as}, \frac{(s-a)(s-b)}{as}, \frac{(s-a)(s-c)}{as}\right)=(a^2:(s-a)(s-b):(s-a)(s-c)).</cmath> | ||
+ | |||
+ | Recalling that the equation of the incircle is <cmath>a^2yz+b^2xz+c^2xy+(x+y+z)[(s-a)^2x+(s-b)^2y+(s-c)^2z]=0.</cmath> We must show that this equation is true for <math>Q'</math>'s values of <math>x,y,z</math>. | ||
+ | |||
+ | Plugging in our values, this means showing that | ||
+ | <cmath>a^2(s-a)^2(s-b)(s-c)+a^2(s-a)[b^2(s-c)+c^2(s-b)]+as[a^2(s-a)^2+(s-a)(s-b)^3+(s-a)(s-c)^3]=0.</cmath> | ||
+ | Dividing by <math>a(s-a)</math>, this is just | ||
+ | <cmath>a(s-a)(s-b)(s-c)+a[b^2(s-c)+c^2(s-b)]+a^2s(s-a)+s(s-b)^3+s(s-c)^3=0.</cmath> | ||
+ | |||
+ | Plugging in the value of <math>s:</math> | ||
+ | <cmath>\frac{a(-a+b+c)(a-b+c)(a+b-c)}{8}+\frac{ab^2(a+b-c)}{2}+\frac{ac^2(a-b+c)}{2}+\frac{a^2(a+b+c)(-a+b+c)}{4}+\frac{(a+b+c)(a-b+c)^3}{16}+\frac{(a+b+c)(a+b-c)^3}{16}=0.</cmath> | ||
+ | <cmath>2a(-a+b+c)(a-b+c)(a+b-c)+8ab^2(a+b-c)+8ac^2(a-b+c)+4a^2(a+b+c)(-a+b+c)+(a+b+c)(a-b+c)^3+(a+b+c)(a+b-c)^3=0</cmath> | ||
+ | <cmath>2a[(-a+b+c)(a-b+c)(a+b-c)+4b^2(a+b-c)+4c^2(a-b+c)]+(a+b+c)[4a^2(-a+b+c)+(a-b+c)^3+(a+b-c)^3]=0</cmath> | ||
+ | The first bracket is just | ||
+ | <cmath>-a^3-b^3-c^3+a^2b+a^2c+b^2c+b^2a+c^2a+c^2b-2abc+4ab^2+4b^3-4b^2c+4ac^2-4bc^2+4c^3</cmath> | ||
+ | <cmath>=-a^3+3b^3+3c^3+a^2b+a^2c-3b^2c+5b^2a+5c^2a-3c^2b-2abc</cmath> | ||
+ | and the second bracket is <cmath>-2a^3+4a^2b+4a^2c+6ab^2+6ac^2-12abc.</cmath> | ||
+ | Dividing everything by <math>2a</math> gives | ||
+ | <cmath>-a^3+3b^3+3c^3+a^2b+a^2c-3b^2c+5b^2a+5c^2a-3c^2b-2abc+(a+b+c)(-a^2+2ab+2ac+3b^2+3c^2-6bc),</cmath> which is <math>0</math>, as desired. | ||
+ | |||
+ | As <math>Q'</math> lies on the incircle and <math>AD_2</math>, <math>Q'=Q</math>, and our proof is complete. | ||
+ | |||
+ | |||
+ | === Solution 4 === | ||
+ | We again use Barycentric coordinates. As before, let <math>A</math> be <math>(1,0,0)</math>, <math>B</math> be <math>(0,1,0)</math>, and <math>C</math> be <math>(0,0,1)</math>. Also | ||
+ | |||
+ | <cmath>D_{1}=\left( 0,\frac{s-c}{a},\frac{s-b}{a} \right), D_2=\left(0,\frac{s-b}{a},\frac{s-c}{a}\right), E_2=\left(\frac{s-a}{b},0,\frac{s-c}{b}\right), P=\left(\frac{s-a}{s},\frac{s-b}{s},\frac{s-c}{s}\right).</cmath> | ||
+ | |||
+ | Now, consider a point <math>Q'</math> for which <math>AQ'=PD_{2}</math>. Then working component-vise, we get <math>Q'+P=A+D_{2}</math> from which we can easily get the coordinates of <math>Q'</math> as; | ||
+ | |||
+ | <cmath>Q'=\left(\frac{a}{s},\frac{(s-a)(s-b)}{sa},\frac{(s-a)(s-c)}{sa}\right)</cmath> | ||
+ | |||
+ | It suffices to show that <math>Q'=Q</math>. | ||
+ | |||
+ | Let <math>I=\left(\frac{a}{2s},\frac{b}{2s},\frac{c}{2s}\right)</math> be the incenter of triangle <math>ABC</math>. We claim that <math>I</math> is the midpoint of <math>Q'D_{1}</math>. Indeed, | ||
+ | |||
+ | <cmath>\frac{1}{2}\left(0+\frac{a}{s}\right)=\frac{a}{2s}</cmath> | ||
+ | |||
+ | <cmath>\frac{1}{2}\left(\frac{(s-a)(s-b)}{sa}+\frac{s-c}{a}\right)=\frac{(s-a)(s-b)+s(s-c)}{2sa}=\frac{ab}{2sa}=\frac{b}{2s}</cmath> | ||
+ | |||
+ | <cmath>\frac{1}{2}\left(\frac{(s-a)(s-c)}{sa}+\frac{s-b}{a}\right)=\frac{c}{2s}</cmath> | ||
+ | |||
+ | Hence the claim has been proved. | ||
+ | |||
+ | Since <math>I</math> is the center of <math>\omega</math> and the midpoint of <math>Q'D_{1}</math>, thus <math>Q'</math> is the point diametrically opposite to <math>D_{1}</math>, and hence it lies on <math>\omega</math> and closer to <math>A</math> off the 2 points. Thus <math>Q=Q'</math> as desired. | ||
== See also == | == See also == |
Latest revision as of 08:35, 16 November 2017
Contents
Problem
Let be a triangle and let
be its incircle. Denote by
and
the points where
is tangent to sides
and
, respectively. Denote by
and
the points on sides
and
, respectively, such that
and
, and denote by
the point of intersection of segments
and
. Circle
intersects segment
at two points, the closer of which to the vertex
is denoted by
. Prove that
.
Solution
Solution 1
It is well known that the excircle opposite is tangent to
at the point
. (Proof: let the points of tangency of the excircle with the lines
be
respectively. Then
. It follows that
, and
, so
.)
Now consider the homothety that carries the incircle of to its excircle. The homothety also carries
to
(since
are collinear), and carries the tangency points
to
. It follows that
.
By Menelaus' Theorem on
with segment
, it follows that
. It easily follows that
.
Solution 2
The key observation is the following lemma.
Lemma: Segment is a diameter of circle
.
Proof: Let be the center of circle
, i.e.,
is the incenter of triangle
. Extend segment
through
to intersect circle
again at
, and extend segment
through
to intersect segment
at
. We show that
, which in turn implies that
, that is,
is a diameter of
.
Let be the line tangent to circle
at
, and let
intersect the segments
and
at
and
, respectively. Then
is an excircle of triangle
. Let
denote the dilation with its center at
and ratio
. Since
and
,
. Hence
. Thus
,
, and
. It also follows that an excircle
of triangle
is tangent to the side
at
.
It is well known that We compute
. Let
and
denote the points of tangency of circle
with rays
and
, respectively. Then by equal tangents,
,
, and
. Hence
It follows that
Combining these two equations yields
. Thus
that is,
, as desired.
Now we prove our main result. Let and
be the respective midpoints of segments
and
. Then
is also the midpoint of segment
, from which it follows that
is the midline of triangle
. Hence
and
. Similarly, we can prove that
.
2001usamo2-2.png
Let be the centroid of triangle
. Thus segments
and
intersect at
. Define transformation
as the dilation with its center at
and ratio
. Then
and
. Under the dilation, parallel lines go to parallel lines and the intersection of two lines goes to the intersection of their images. Since
and
,
maps lines
and
to lines
and
, respectively. It also follows that
and
or
This yields
as desired.
Note: We used directed lengths in our calculations to avoid possible complications caused by the different shapes of triangle .
Solution 3
Here is a rather nice solution using barycentric coordinates:
Let be
,
be
, and
be
. Let the side lengths of the triangle be
and the semi-perimeter
.
Now, Thus,
Therefore, and
Clearly then, the non-normalized coordinates of
Normalizing, we have that
Now, we find the point inside the triangle on the line
such that
. It is then sufficient to show that this point lies on the incircle.
is the fraction
of the way "up" the line segment from
to
. Thus, we are looking for the point that is
of the way "down" the line segment from
to
, or, the fraction
of the way "up".
Thus, has normalized
-coordinate
.
As the line has equation
, it can easily be found that
Recalling that the equation of the incircle is We must show that this equation is true for
's values of
.
Plugging in our values, this means showing that
Dividing by
, this is just
Plugging in the value of
The first bracket is just
and the second bracket is
Dividing everything by
gives
which is
, as desired.
As lies on the incircle and
,
, and our proof is complete.
Solution 4
We again use Barycentric coordinates. As before, let be
,
be
, and
be
. Also
Now, consider a point for which
. Then working component-vise, we get
from which we can easily get the coordinates of
as;
It suffices to show that .
Let be the incenter of triangle
. We claim that
is the midpoint of
. Indeed,
Hence the claim has been proved.
Since is the center of
and the midpoint of
, thus
is the point diametrically opposite to
, and hence it lies on
and closer to
off the 2 points. Thus
as desired.
See also
2001 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.