Difference between revisions of "2008 USAMO Problems/Problem 2"
(4 intermediate revisions by 2 users not shown) | |||
Line 5: | Line 5: | ||
=== Solution 1 (synthetic) === | === Solution 1 (synthetic) === | ||
+ | Without Loss of Generality, assume <math>AB >AC</math>. It is sufficient to prove that <math>\angle OFA = 90^{\circ}</math>, as this would immediately prove that <math>A,P,O,F,N</math> are concyclic. | ||
+ | By applying the Menelaus' Theorem in the Triangle <math>\triangle BFC</math> for the transversal <math>E,M,D</math>, we have (in magnitude) | ||
+ | <cmath> \frac{FE}{EC} \cdot \frac{CM}{MB} \cdot \frac{BD}{DF} = 1 \iff \frac{FE}{EC} = \frac{DF}{BD} | ||
+ | </cmath> | ||
+ | Here, we used that <math>BM=MC</math>, as <math>M</math> is the midpoint of <math>BC</math>. Now, since <math>EC =EA</math> and <math>BD=DA</math>, we have | ||
+ | <cmath> \frac{FE}{EA} = \frac{DF}{DA} \iff \frac{DA}{AE} = \frac{DF}{FE} \iff AF \text{ bisects exterior } \angle EFD | ||
+ | </cmath> | ||
+ | Now, note that <math>OE</math> bisects the exterior <math>\angle FED</math> and <math>OD</math> bisects exterior <math>\angle FDE</math>, making <math>O</math> the <math>F</math>-excentre of <math>\triangle FED</math>. This implies that <math>OF</math> bisects interior <math>\angle EFD</math>, making <math>OF \perp AF</math>, as was required. | ||
+ | |||
+ | === Solution 2 (complex) === | ||
+ | |||
+ | Let <math>A=1,B=b,C=c</math> where <math>b,c</math> all lie on the unit circle. Then <math>O</math> is 0. As noted earlier, <math>(FOBC)</math> is cyclic. We will find the ghost point <math>F',</math> the second intersection of <math>OBC</math> and <math>ANP</math>. | ||
+ | |||
+ | We know that these two circles already intersect at <math>O</math> so we can reflect over the line between their centers. The center of <math>ANPO</math> is the midpoint of <math>AO</math> namely <math>\frac12</math>. With the tangent formula and then taking the midpoint, we find that the center of <math>OBC</math> is <math>\frac{bc}{b+c}.</math> Then we want to find the reflection of 0 over the line through <math>\frac12</math> and <math>\frac{bc}{b+c}.</math> Then we get | ||
+ | |||
+ | <cmath> \begin{align*} | ||
+ | f' &= \frac{\left(\frac{bc}{b+c}-\overline{\frac{bc}{b+c}}\right)\div2}{(1/2)-\overline{\frac{bc}{b+c}}}\\ | ||
+ | &= \frac{\frac{bc-1}{2(b+c)}}{\frac{b+c-2}{2(b+c)}}\\ | ||
+ | &= \frac{bc-1}{b+c-2}. | ||
+ | \end{align*} </cmath> | ||
+ | |||
+ | Now it remains to show <math>\angle F'BA=\angle ABM;</math> the other angle equality would follow by symmetry. | ||
+ | |||
+ | Then we get: | ||
+ | <cmath> \begin{align*} | ||
+ | \frac{f'-b}{b-a}\div\frac{b-a}{a-m} | ||
+ | &=\frac{\frac{bc-1}{b+c-2}-b}{b-1}\div\frac{b-1}{1-\frac{b+c}2}\\ | ||
+ | &=\frac{\frac{bc-1-b(b+c-2)}{b+c-2}(2-b+c)}{2(b-1)^2}\\ | ||
+ | &=\frac{b^2-2b+1}{2(b-1)^2}\\ | ||
+ | &=\frac12. | ||
+ | \end{align*} </cmath> | ||
+ | Thus <math>\measuredangle F'BA=\measuredangle BAM,</math> so <math>F'=F</math> and we're done. | ||
+ | |||
+ | ~cocohearts | ||
+ | |||
+ | === Solution 3 (synthetic) === | ||
<center><asy> | <center><asy> | ||
/* setup and variables */ | /* setup and variables */ | ||
Line 78: | Line 114: | ||
<cmath>\angle EMO = \angle DMO = \angle DPF = \angle OPF</cmath> | <cmath>\angle EMO = \angle DMO = \angle DPF = \angle OPF</cmath> | ||
Hence <math>\angle OPF = \angle ONF</math>, so <math>FONP</math> is cyclic. In other words, <math>F</math> lies on the circumcircle of <math>\triangle PON</math>. Note that <math>\angle ONA = \angle OPA = 90</math>, so <math>APON</math> is cyclic. In other words, <math>A</math> lies on the circumcircle of <math>\triangle PON</math>. <math>A</math>, <math>P</math>, <math>N</math>, <math>O</math>, and <math>F</math> all lie on the circumcircle of <math>\triangle PON</math>. Hence <math>A</math>, <math>P</math>, <math>F</math>, and <math>N</math> lie on a circle, as desired. | Hence <math>\angle OPF = \angle ONF</math>, so <math>FONP</math> is cyclic. In other words, <math>F</math> lies on the circumcircle of <math>\triangle PON</math>. Note that <math>\angle ONA = \angle OPA = 90</math>, so <math>APON</math> is cyclic. In other words, <math>A</math> lies on the circumcircle of <math>\triangle PON</math>. <math>A</math>, <math>P</math>, <math>N</math>, <math>O</math>, and <math>F</math> all lie on the circumcircle of <math>\triangle PON</math>. Hence <math>A</math>, <math>P</math>, <math>F</math>, and <math>N</math> lie on a circle, as desired. | ||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
=== Solution 4 (synthetic) === | === Solution 4 (synthetic) === | ||
Line 105: | Line 126: | ||
It is easy to see <math>Q</math> (the intersection of ray <math>OM</math> and the circumcircle of <math>\triangle BOC</math>) is colinear with <math>A</math> and <math>F'</math>, and because line <math>OM</math> is the diameter of that circle, <math>\angle QBO = \angle QCO = 90</math>, so <math>Q</math> is the point <math>Q</math> in the lemma; hence, we may apply the lemma. From here, it is simple angle-chasing to show that <math>F'</math> satisfies the original construction for <math>F</math>, showing <math>F=F'</math>; we are done. | It is easy to see <math>Q</math> (the intersection of ray <math>OM</math> and the circumcircle of <math>\triangle BOC</math>) is colinear with <math>A</math> and <math>F'</math>, and because line <math>OM</math> is the diameter of that circle, <math>\angle QBO = \angle QCO = 90</math>, so <math>Q</math> is the point <math>Q</math> in the lemma; hence, we may apply the lemma. From here, it is simple angle-chasing to show that <math>F'</math> satisfies the original construction for <math>F</math>, showing <math>F=F'</math>; we are done. | ||
− | + | ||
=== Solution 5 (trigonometric) === | === Solution 5 (trigonometric) === | ||
Line 150: | Line 171: | ||
Then <math>FP</math> is a median, use the lemma again to show that <math>AFP = C</math>, and similarly <math>AFN = B</math>, so you're done. | Then <math>FP</math> is a median, use the lemma again to show that <math>AFP = C</math>, and similarly <math>AFN = B</math>, so you're done. | ||
− | + | ||
=== Solution 8 (inversion) (Official Solution #2) === | === Solution 8 (inversion) (Official Solution #2) === | ||
Line 189: | Line 210: | ||
<cmath>\angle AFO = \angle AFB - \angle OFB = (180^\circ - \angle BAC) - (90^\circ - \angle BAC) = 90^\circ.</cmath> | <cmath>\angle AFO = \angle AFB - \angle OFB = (180^\circ - \angle BAC) - (90^\circ - \angle BAC) = 90^\circ.</cmath> | ||
This completes the proof. | This completes the proof. | ||
+ | === Solution 10 (Anti-Steiner point) === | ||
+ | First, let <math>O</math> be the circumcenter of <math>\triangle{ABC}</math>. Note that since <math>AP \perp PO</math> and <math>AN \perp NO</math>, then <math>A</math>, <math>N</math>, <math>O</math> and <math>P</math>, are concyclic. | ||
+ | |||
+ | Notice that <math>NM \parallel AB</math> and <math>AB \perp PO</math>, which means <math>NM \perp PO</math>. It follows analogously that <math>PM \perp NO</math>. This means that <math>M</math> is the orthocenter of triangle <math>{NOP}</math>. | ||
+ | |||
+ | Now consider what happens when we reflect line <math>AM</math> over line <math>PO</math>. Since <math>PO</math> is just the perpendicular bisector of <math>AB</math>, point <math>A</math> will map to point <math>B</math>, and point <math>D</math>, which is both line <math>AM</math> and <math>PO</math>, will map to itself. Therefore, we get line <math>BD</math> from this reflection. It follows analogously that by reflecting line <math>AM</math> over line <math>NO</math>, we get <math>CE</math>. | ||
+ | |||
+ | Since line <math>AM</math> passes through <math>M</math>, the orthocenter of triangle <math>{NOP}</math>, its reflections over sides <math>PO</math> and <math>NO</math>, which correspond to <math>BD</math> and <math>CE</math> respectively, intersect on <math>\odot{NOP}</math> (this is the Anti-Steiner point application). Therefore, <math>F</math>, the intersection of <math>BD</math> and <math>CE</math>, lies on <math>\odot{NOP}</math>. | ||
+ | |||
+ | Since <math>A</math>, <math>N</math>, <math>O</math>, and <math>P</math> are concyclic, then <math>\odot{NOP}</math> is the same as <math>\odot{ANP}</math>, so <math>F</math> lies on <math>\odot{ANP}</math>, which is the same as saying <math>A</math>, <math>N</math>, <math>F</math>, and <math>P</math> are concyclic, as desired. | ||
+ | |||
Line 198: | Line 230: | ||
{{USAMO newbox|year=2008|num-b=1|num-a=3}} | {{USAMO newbox|year=2008|num-b=1|num-a=3}} | ||
− | |||
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 11:54, 1 May 2024
Contents
- 1 Problem
- 2 Solutions
- 2.1 Solution 1 (synthetic)
- 2.2 Solution 2 (complex)
- 2.3 Solution 3 (synthetic)
- 2.4 Solution 4 (synthetic)
- 2.5 Solution 5 (trigonometric)
- 2.6 Solution 6 (isogonal conjugates)
- 2.7 Solution 7 (symmedians)
- 2.8 Solution 8 (inversion) (Official Solution #2)
- 2.9 Solution 9 (Official Solution #1)
- 2.10 Solution 10 (Anti-Steiner point)
- 3 See Also
Problem
(Zuming Feng) Let be an acute, scalene triangle, and let
,
, and
be the midpoints of
,
, and
, respectively. Let the perpendicular bisectors of
and
intersect ray
in points
and
respectively, and let lines
and
intersect in point
, inside of triangle
. Prove that points
,
,
, and
all lie on one circle.
Solutions
Solution 1 (synthetic)
Without Loss of Generality, assume . It is sufficient to prove that
, as this would immediately prove that
are concyclic.
By applying the Menelaus' Theorem in the Triangle
for the transversal
, we have (in magnitude)
Here, we used that
, as
is the midpoint of
. Now, since
and
, we have
Now, note that
bisects the exterior
and
bisects exterior
, making
the
-excentre of
. This implies that
bisects interior
, making
, as was required.
Solution 2 (complex)
Let where
all lie on the unit circle. Then
is 0. As noted earlier,
is cyclic. We will find the ghost point
the second intersection of
and
.
We know that these two circles already intersect at so we can reflect over the line between their centers. The center of
is the midpoint of
namely
. With the tangent formula and then taking the midpoint, we find that the center of
is
Then we want to find the reflection of 0 over the line through
and
Then we get
Now it remains to show the other angle equality would follow by symmetry.
Then we get:
Thus
so
and we're done.
~cocohearts
Solution 3 (synthetic)
![[asy] /* setup and variables */ size(280); pathpen = black + linewidth(0.7); pointpen = black; pen s = fontsize(8); pair B=(0,0),C=(5,0),A=(1,4); /* A.x > C.x/2 */ /* construction and drawing */ pair P=(A+B)/2,M=(B+C)/2,N=(A+C)/2,D=IP(A--M,P--P+5*(P-bisectorpoint(A,B))),E=IP(A--M,N--N+5*(bisectorpoint(A,C)-N)),F=IP(B--B+5*(D-B),C--C+5*(E-C)),O=circumcenter(A,B,C); D(MP("A",A,(0,1),s)--MP("B",B,SW,s)--MP("C",C,SE,s)--A--MP("M",M,s)); D(B--D(MP("D",D,NE,s))--MP("P",P,(-1,0),s)--D(MP("O",O,(0,1),s))); D(D(MP("E",E,SW,s))--MP("N",N,(1,0),s)); D(C--D(MP("F",F,NW,s))); D(B--O--C,linetype("4 4")+linewidth(0.7)); D(M--N,linetype("4 4")+linewidth(0.7)); D(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5)); D(anglemark(B,A,C)); MP("y",A,(0,-6));MP("z",A,(4,-6)); D(anglemark(B,F,C,4),linewidth(0.6));D(anglemark(B,O,C,4),linewidth(0.6)); picture p = new picture; draw(p,circumcircle(B,O,C),linetype("1 4")+linewidth(0.7)); clip(p,B+(-5,0)--B+(-5,A.y+2)--C+(5,A.y+2)--C+(5,0)--cycle); add(p); /* D(circumcircle(A,P,N),linetype("4 4")+linewidth(0.7)); */ [/asy]](http://latex.artofproblemsolving.com/5/9/c/59c1afd86c69597587b940e9645c56d85342ba62.png)
Without loss of generality . The intersection of
and
is
, the circumcenter of
.
Let and
. Note
lies on the perpendicular bisector of
, so
. So
. Similarly,
, so
. Notice that
intercepts the minor arc
in the circumcircle of
, which is double
. Hence
, so
is cyclic.
Lemma. is directly similar to
Proof.
since
,
,
are collinear,
is cyclic, and
. Also
because
, and
is the medial triangle of
so
. Hence
.
Notice that since
.
. Then
Hence
.
Hence is similar to
by AA similarity. It is easy to see that they are oriented such that they are directly similar.
End Lemma
![[asy] /* setup and variables */ size(280); pathpen = black + linewidth(0.7); pointpen = black; pen s = fontsize(8); pair B=(0,0),C=(5,0),A=(1,4); /* A.x > C.x/2 */ /* construction and drawing */ pair P=(A+B)/2,M=(B+C)/2,N=(A+C)/2,D=IP(A--M,P--P+5*(P-bisectorpoint(A,B))),E=IP(A--M,N--N+5*(bisectorpoint(A,C)-N)),F=IP(B--B+5*(D-B),C--C+5*(E-C)),O=circumcenter(A,B,C); D(MP("A",A,(0,1),s)--MP("B",B,SW,s)--MP("C",C,SE,s)--A--MP("M",M,s)); D(B--D(MP("D",D,NE,s))--MP("P",P,(-1,0),s)--D(MP("O",O,(1,0),s))); D(D(MP("E",E,SW,s))--MP("N",N,(1,0),s)); D(C--D(MP("F",F,NW,s))); D(B--O--C,linetype("4 4")+linewidth(0.7)); D(F--N); D(O--M); D(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5)); /* commented in above asy D(circumcircle(A,P,N),linetype("4 4")+linewidth(0.7)); D(anglemark(B,A,C)); MP("y",A,(0,-6));MP("z",A,(4,-6)); D(anglemark(B,F,C,4),linewidth(0.6));D(anglemark(B,O,C,4),linewidth(0.6)); picture p = new picture; draw(p,circumcircle(B,O,C),linetype("1 4")+linewidth(0.7)); clip(p,B+(-5,0)--B+(-5,A.y+2)--C+(5,A.y+2)--C+(5,0)--cycle); add(p); */ [/asy]](http://latex.artofproblemsolving.com/f/2/1/f21402ee04f8bab4de4509f8e2cdcae9d4f26c5a.png)
By the similarity in the Lemma, .
so
by SAS similarity. Hence
Using essentially the same angle chasing, we can show that
is directly similar to
. It follows that
is directly similar to
. So
Hence
, so
is cyclic. In other words,
lies on the circumcircle of
. Note that
, so
is cyclic. In other words,
lies on the circumcircle of
.
,
,
,
, and
all lie on the circumcircle of
. Hence
,
,
, and
lie on a circle, as desired.
Solution 4 (synthetic)
This solution utilizes the phantom point method. Clearly, APON are cyclic because . Let the circumcircles of triangles
and
intersect at
and
.
Lemma. If are points on circle
with center
, and the tangents to
at
intersect at
, then
is the symmedian from
to
.
Proof. This is fairly easy to prove (as H, O are isogonal conjugates, plus using SAS similarity), but the author lacks time to write it up fully, and will do so soon.
End Lemma
It is easy to see (the intersection of ray
and the circumcircle of
) is colinear with
and
, and because line
is the diameter of that circle,
, so
is the point
in the lemma; hence, we may apply the lemma. From here, it is simple angle-chasing to show that
satisfies the original construction for
, showing
; we are done.
Solution 5 (trigonometric)
By the Law of Sines, . Since
and similarly
, we cancel to get
. Obviously,
so
.
Then and
. Subtracting these two equations,
so
. Therefore,
(by AA similarity), so a spiral similarity centered at
takes
to
and
to
. Therefore, it takes the midpoint of
to the midpoint of
, or
to
. So
and
is cyclic.
Solution 6 (isogonal conjugates)
![[asy] /* setup and variables */ size(280); pathpen = black + linewidth(0.7); pointpen = black; pen s = fontsize(8); pair B=(0,0),C=(5,0),A=(4,4); /* A.x > C.x/2 */ /* construction and drawing */ pair P=(A+B)/2,M=(B+C)/2,N=(A+C)/2,D=IP(A--M,P--P+5*(P-bisectorpoint(A,B))),E=IP(A--M,N--N+5*(bisectorpoint(A,C)-N)),F=IP(B--B+5*(D-B),C--C+5*(E-C)),O=circumcenter(A,B,C); D(MP("A",A,(0,1),s)--MP("B",B,SW,s)--MP("C",C,SE,s)--A--MP("M",M,s)); D(C--D(MP("E",E,NW,s))--MP("N",N,(1,0),s)--D(MP("O",O,SW,s))); D(D(MP("D",D,SE,s))--MP("P",P,W,s)); D(B--D(MP("F",F,s))); D(O--A--F,linetype("4 4")+linewidth(0.7)); D(MP("O'",circumcenter(A,P,N),NW,s)); D(circumcircle(A,P,N),linetype("4 4")+linewidth(0.7)); D(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5)); picture p = new picture; draw(p,circumcircle(B,O,C),linetype("1 4")+linewidth(0.7)); draw(p,circumcircle(A,B,C),linetype("1 4")+linewidth(0.7)); clip(p,B+(-5,0)--B+(-5,A.y+2)--C+(5,A.y+2)--C+(5,0)--cycle); add(p); [/asy]](http://latex.artofproblemsolving.com/e/f/4/ef48251bf50ebbfa10db0114d80f3a0f7efe3276.png)
Construct on
such that
. Then
. Then
, so
, or
. Then
, so
. Then we have
and
. So
and
are isogonally conjugate. Thus
. Then
.
If is the circumcenter of
then
so
is cyclic. Then
.
Then . Then
is a right triangle.
Now by the homothety centered at with ratio
,
is taken to
and
is taken to
. Thus
is taken to the circumcenter of
and is the midpoint of
, which is also the circumcenter of
, so
all lie on a circle.
Solution 7 (symmedians)
Median of a triangle
implies
.
Trig ceva for
shows that
is a symmedian.
Then
is a median, use the lemma again to show that
, and similarly
, so you're done.
Solution 8 (inversion) (Official Solution #2)
Invert the figure about a circle centered at , and let
denote the image of the point
under this inversion. Find point
so that
is a parallelogram and let
denote the center of this parallelogram. Note that
and
. Because
is the midpoint of
and
is the midpoint of
, we also have
. Thus
Hence quadrilateral
is cyclic and, by a similar argument, quadrilateral
is also cyclic. Because the images under the inversion of lines
and
are circles that intersect in
and
, it follows that
.
Next note that ,
, and
are collinear and are the images of
,
, and
, respectively, under a homothety centered at
and with ratio
. It follows that
,
, and
are collinear, and then that the points
,
,
, and
lie on a circle.
![2008usamo2-sol8.png](https://wiki-images.artofproblemsolving.com//0/05/2008usamo2-sol8.png)
Solution 9 (Official Solution #1)
Let be the circumcenter of triangle
. We prove that
It will then follow that
lie on the circle with diameter
. Indeed, the fact that the first two angles in
are right is immediate because
and
are the perpendicular bisectors of
and
, respectively. Thus we need only prove that
.
![2008usamo2-sol9.png](https://wiki-images.artofproblemsolving.com//d/df/2008usamo2-sol9.png)
We may assume, without loss of generality, that . This leads to configurations similar to the ones shown above. The proof can be adapted to other configurations. Because
is the perpendicular bisector of
, it follows that triangle
is an isosceles triangle with
. Likewise, triangle
is isosceles with
. Let
and
, so
.
Applying the Law of Sines to triangles and
gives
Taking the quotient of the two equations and noting that
, we find
Because
, we have
Applying the Law of Sines to triangles
and
, we find
Taking the quotient of the two equations yields
so by
,
Because
is an exterior angle to triangle
, we have
. Similarly,
. Hence
Thus
, so
is cyclic. In addition,
and hence, from
,
. Because
is cyclic and
is isosceles with vertex angle
, we have
. Therefore,
This completes the proof.
Solution 10 (Anti-Steiner point)
First, let be the circumcenter of
. Note that since
and
, then
,
,
and
, are concyclic.
Notice that and
, which means
. It follows analogously that
. This means that
is the orthocenter of triangle
.
Now consider what happens when we reflect line over line
. Since
is just the perpendicular bisector of
, point
will map to point
, and point
, which is both line
and
, will map to itself. Therefore, we get line
from this reflection. It follows analogously that by reflecting line
over line
, we get
.
Since line passes through
, the orthocenter of triangle
, its reflections over sides
and
, which correspond to
and
respectively, intersect on
(this is the Anti-Steiner point application). Therefore,
, the intersection of
and
, lies on
.
Since ,
,
, and
are concyclic, then
is the same as
, so
lies on
, which is the same as saying
,
,
, and
are concyclic, as desired.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
- <url>viewtopic.php?t=202907 Discussion on AoPS/MathLinks</url>
2008 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.