2019 USAJMO Problems/Problem 4
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?
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-
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.
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
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 0.92, xmax = 28.12, ymin = -14.66, ymax = 5.16; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902);
/* draw figures */
draw((3.52,1.38)--(5.8,-4.7), linewidth(1) + rvwvcq); draw((5.8,-4.7)--(10.3,-4.9), linewidth(1) + wvvxds); draw((10.3,-4.9)--(3.52,1.38), linewidth(1) + rvwvcq); draw(circle((9.325581455949688,-7.26800412402583), 2.399418339060914), linewidth(1) + sexdts); draw((5.8,-4.7)--(7.0789360558927,-8.1104961490472), linewidth(1) + rvwvcq); draw((10.3,-4.9)--(10.956076160142976,-5.507692962492314), linewidth(1) + rvwvcq); draw((8.503617697888226,-3.2360942688404215)--(6.711506849315068,-7.130684931506849), linewidth(1) + wvvxds);
/* dots and labels */
dot((3.52,1.38),dotstyle); label("", (3.6,1.58), NE * labelscalefactor); dot((5.8,-4.7),dotstyle); label("", (5.88,-4.5), NE * labelscalefactor); dot((10.3,-4.9),dotstyle); label("", (10.38,-4.7), NE * labelscalefactor); dot((9.325581455949688,-7.26800412402583),dotstyle); label("", (9.4,-7.06), NE * labelscalefactor); dot((7.0789360558927,-8.1104961490472),linewidth(4pt) + dotstyle); label("", (7.16,-7.96), NE * labelscalefactor); dot((10.956076160142976,-5.507692962492314),linewidth(4pt) + dotstyle); label("", (11.04,-5.34), NE * labelscalefactor); dot((8.503617697888226,-3.2360942688404215),dotstyle); label("E", (8.58,-3.04), NE * labelscalefactor); dot((6.711506849315068,-7.130684931506849),dotstyle); label("", (6.8,-6.94), NE * labelscalefactor); dot((7.139200885553699,-6.201226130819783),dotstyle); label("", (7.22,-6), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy] 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
|2019 USAJMO (Problems • Resources)|
|1 • 2 • 3 • 4 • 5 • 6|
|All USAJMO Problems and Solutions|