Difference between revisions of "2019 USAJMO Problems/Problem 4"
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<math>(*)</math> Let <math>ABC</math> be a triangle with <math>\angle ABC</math> obtuse. The <math>A</math>''-excircle'' is a circle in the exterior of <math>\triangle ABC</math> that is tangent to side <math>\overline{BC}</math> of the triangle and tangent to the extensions of the other two sides. Let <math>E</math>, <math>F</math> be the feet of the altitudes from <math>B</math> and <math>C</math> to lines <math>AC</math> and <math>AB</math>, respectively. Can line <math>EF</math> be tangent to the <math>A</math>-excircle? | <math>(*)</math> Let <math>ABC</math> be a triangle with <math>\angle ABC</math> obtuse. The <math>A</math>''-excircle'' is a circle in the exterior of <math>\triangle ABC</math> that is tangent to side <math>\overline{BC}</math> of the triangle and tangent to the extensions of the other two sides. Let <math>E</math>, <math>F</math> be the feet of the altitudes from <math>B</math> and <math>C</math> to lines <math>AC</math> and <math>AB</math>, respectively. Can line <math>EF</math> be tangent to the <math>A</math>-excircle? | ||
− | ==Solution== | + | ==Solution 1== |
Instead of trying to find a synthetic way to describe <math>EF</math> being tangent to the <math>A</math>-excircle (very hard), we instead consider the foot of the perpendicular from the <math>A</math>-excircle to <math>EF</math>, hoping to force something via the length of the perpendicular. It would be nice if there were an easier way to describe <math>EF</math>, something more closely related to the <math>A</math>-excircle; as we are considering perpendicularity, if we could generate a line parallel to <math>EF</math>, that would be good. | Instead of trying to find a synthetic way to describe <math>EF</math> being tangent to the <math>A</math>-excircle (very hard), we instead consider the foot of the perpendicular from the <math>A</math>-excircle to <math>EF</math>, hoping to force something via the length of the perpendicular. It would be nice if there were an easier way to describe <math>EF</math>, something more closely related to the <math>A</math>-excircle; as we are considering perpendicularity, if we could generate a line parallel to <math>EF</math>, that would be good. | ||
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==Solution 4== | ==Solution 4== | ||
− | + | <asy> | |
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ | /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ | ||
import graph; size(10cm); | import graph; size(10cm); | ||
Line 49: | Line 49: | ||
/* dots and labels */ | /* dots and labels */ | ||
dot((3.52,1.38),dotstyle); | dot((3.52,1.38),dotstyle); | ||
− | label(" | + | label("$A$", (3.6,1.58), NE * labelscalefactor); |
dot((5.8,-4.7),dotstyle); | dot((5.8,-4.7),dotstyle); | ||
− | label(" | + | label("$B$", (5.88,-4.5), NE * labelscalefactor); |
dot((10.3,-4.9),dotstyle); | dot((10.3,-4.9),dotstyle); | ||
− | label(" | + | label("$C$", (10.38,-4.7), NE * labelscalefactor); |
dot((9.325581455949688,-7.26800412402583),dotstyle); | dot((9.325581455949688,-7.26800412402583),dotstyle); | ||
− | label(" | + | label("$I_a$", (9.4,-7.06), NE * labelscalefactor); |
dot((7.0789360558927,-8.1104961490472),linewidth(4pt) + dotstyle); | dot((7.0789360558927,-8.1104961490472),linewidth(4pt) + dotstyle); | ||
− | label(" | + | label("$G$", (7.16,-7.96), NE * labelscalefactor); |
dot((10.956076160142976,-5.507692962492314),linewidth(4pt) + dotstyle); | dot((10.956076160142976,-5.507692962492314),linewidth(4pt) + dotstyle); | ||
− | label(" | + | label("$H$", (11.04,-5.34), NE * labelscalefactor); |
dot((8.503617697888226,-3.2360942688404215),dotstyle); | dot((8.503617697888226,-3.2360942688404215),dotstyle); | ||
label("E", (8.58,-3.04), NE * labelscalefactor); | label("E", (8.58,-3.04), NE * labelscalefactor); | ||
dot((6.711506849315068,-7.130684931506849),dotstyle); | dot((6.711506849315068,-7.130684931506849),dotstyle); | ||
− | label(" | + | label("$F$", (6.8,-6.94), NE * labelscalefactor); |
dot((7.139200885553699,-6.201226130819783),dotstyle); | dot((7.139200885553699,-6.201226130819783),dotstyle); | ||
− | label(" | + | label("$X$", (7.22,-6), NE * labelscalefactor); |
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | ||
/* end of picture */ | /* end of picture */ | ||
− | + | </asy> | |
We claim that the answer is no. We proceed with contradiction. Suppose that <math>EF</math> is indeed tangent to the a-excenter. Define the point of tangency to be <math>X</math>. Let <math>G</math> to be the intersection of the a-excircle with the extension of <math>AB</math> and <math>H</math> to be the intersection of the a-excircle with the extension of <math>AC</math>. Define <math>I_a</math> to be the a-excenter. It is a well-known fact that <math>I_a</math> lies on the angle bisector of <math>\angle BAC</math>. | We claim that the answer is no. We proceed with contradiction. Suppose that <math>EF</math> is indeed tangent to the a-excenter. Define the point of tangency to be <math>X</math>. Let <math>G</math> to be the intersection of the a-excircle with the extension of <math>AB</math> and <math>H</math> to be the intersection of the a-excircle with the extension of <math>AC</math>. Define <math>I_a</math> to be the a-excenter. It is a well-known fact that <math>I_a</math> lies on the angle bisector of <math>\angle BAC</math>. | ||
For convenience, let <math>AB = c, AC = b, BC = a</math>, <math>\angle CAB = A, \angle CBA = B, \angle ACB = C</math>. | For convenience, let <math>AB = c, AC = b, BC = a</math>, <math>\angle CAB = A, \angle CBA = B, \angle ACB = C</math>. |
Latest revision as of 19:27, 12 April 2021
Problem
Let be a triangle with obtuse. The -excircle is a circle in the exterior of that is tangent to side of the triangle and tangent to the extensions of the other two sides. Let , be the feet of the altitudes from and to lines and , respectively. Can line be tangent to the -excircle?
Solution 1
Instead of trying to find a synthetic way to describe being tangent to the -excircle (very hard), we instead consider the foot of the perpendicular from the -excircle to , hoping to force something via the length of the perpendicular. It would be nice if there were an easier way to describe , something more closely related to the -excircle; as we are considering perpendicularity, if we could generate a line parallel to , that would be good.
So we recall that it is well known that triangle is similar to . This motivates reflecting over the angle bisector at to obtain , which is parallel to for obvious reasons.
Furthermore, as reflection preserves intersection, is tangent to the reflection of the -excircle over the -angle bisector. But it is well-known that the -excenter lies on the -angle bisector, so the -excircle must be preserved under reflection over the -excircle. Thus is tangent to the -excircle.Yet for all lines parallel to , there are only two lines tangent to the -excircle, and only one possibility for , so .
Thus as is isoceles, contradiction. -alifenix-
Solution 2
The answer is no.
Suppose otherwise. Consider the reflection over the bisector of . This swaps rays and ; suppose and are sent to and . Note that the -excircle is fixed, so line must also be tangent to the -excircle.
Since is cyclic, we obtain , so . However, as is a chord in the circle with diameter , .
If then too, so then lies inside and cannot be tangent to the excircle.
The remaining case is when . In this case, is also a diameter, so is a rectangle. In particular . However, by the existence of the orthocenter, the lines and must intersect, contradiction.
Solution 3
The answer is .
Suppose for the sake of contradiction that it is possible for to be tangent to the -excircle. Call the tangency point , and let denote the contact points of with the -excircle, respectively. Let denote the semiperimeter of . By equal tangents, we have It is also well known that , so It is well known (by an easy angle chase) that , so we must have the ratio of similitude is . In particular, This results in which is absurd since is a right triangle. We reached a contradiction, so we are done. ~ Mathscienceclass
Solution 4
We claim that the answer is no. We proceed with contradiction. Suppose that is indeed tangent to the a-excenter. Define the point of tangency to be . Let to be the intersection of the a-excircle with the extension of and to be the intersection of the a-excircle with the extension of . Define to be the a-excenter. It is a well-known fact that lies on the angle bisector of . For convenience, let , . Notice that by Power of a Point: Therefore, adding, we see that: It would be rather nice if we could re-write . Indeed, we can: and similarly for : We now seek to re-write . Using the law of cosines: Therefore, Putting this all together, we see that: Now, we seek to write in terms of the side lengths of the triangle. Notice that: where is the semi-perimeter. We get that: Using the fact that , we have that: Therefore, Returning to our original problem: It is a well-known fact that , so: which implies that: which is a contradiction. Hence, our original assumption that is tangent to the a-excircle is incorrect. ~AopsUser101
See also
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
2019 USAJMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |