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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The remaining case is when <math>EF = BC</math>. In this case, <math>\overline{EF}</math> is also a diameter, so <math>BECF</math> is a rectangle. In particular <math>\overline{BE} \parallel \overline{CF}</math>. However, by the existence of the orthocenter, the lines <math>BE</math> and <math>CF</math> must intersect, contradiction. | The remaining case is when <math>EF = BC</math>. In this case, <math>\overline{EF}</math> is also a diameter, so <math>BECF</math> is a rectangle. In particular <math>\overline{BE} \parallel \overline{CF}</math>. However, by the existence of the orthocenter, the lines <math>BE</math> and <math>CF</math> must intersect, contradiction. | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | The answer is <math>\boxed{\text{no}}</math>. | ||
+ | |||
+ | Suppose for the sake of contradiction that it is possible for <math>EF</math> to be tangent to the <math>A</math>-excircle. Call the tangency point <math>T</math>, and let <math>S_1, S_2</math> denote the contact points of <math>AB, AC</math> with the <math>A</math>-excircle, respectively. Let <math>s</math> denote the semiperimeter of <math>ABC</math>. By equal tangents, we have <cmath>ET = ES_2, FT = FS_1 \implies EF = ES_2+FS_2</cmath>It is also well known that <math>AS_1 = AS_2 = \frac{s}{2}</math>, so <cmath>EF = ES_2+FS_2 = (AS_2-AE)+(AS_1-AF) = s-AE-AF \implies s=AE+AF+EF</cmath>It is well known (by an easy angle chase) that <math>\triangle AEF \sim \triangle ABC</math>, so we must have the ratio of similitude is <math>2</math>. In particular, <cmath>AB=2 \cdot AE, AC=2 \cdot AF</cmath>This results in <cmath>\angle ABE = 30^{\circ}, \angle CBF = 30^{\circ} \implies \angle EBC = 120^{\circ}</cmath>which is absurd since <math>\triangle BEC</math> is a right triangle. We reached a contradiction, so we are done. <math>\blacksquare</math> ~ Mathscienceclass | ||
+ | |||
+ | ==Solution 4== | ||
+ | <asy> | ||
+ | /* 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("$A$", (3.6,1.58), NE * labelscalefactor); | ||
+ | dot((5.8,-4.7),dotstyle); | ||
+ | label("$B$", (5.88,-4.5), NE * labelscalefactor); | ||
+ | dot((10.3,-4.9),dotstyle); | ||
+ | label("$C$", (10.38,-4.7), NE * labelscalefactor); | ||
+ | dot((9.325581455949688,-7.26800412402583),dotstyle); | ||
+ | label("$I_a$", (9.4,-7.06), NE * labelscalefactor); | ||
+ | dot((7.0789360558927,-8.1104961490472),linewidth(4pt) + dotstyle); | ||
+ | label("$G$", (7.16,-7.96), NE * labelscalefactor); | ||
+ | dot((10.956076160142976,-5.507692962492314),linewidth(4pt) + dotstyle); | ||
+ | label("$H$", (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("$F$", (6.8,-6.94), NE * labelscalefactor); | ||
+ | dot((7.139200885553699,-6.201226130819783),dotstyle); | ||
+ | label("$X$", (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 <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>. | ||
+ | Notice that by Power of a Point: | ||
+ | <cmath>FX = FG.</cmath> | ||
+ | <cmath>XE = EH.</cmath> | ||
+ | Therefore, adding, we see that: | ||
+ | <cmath>FE = FX + XE = FG + EH.</cmath> | ||
+ | It would be rather nice if we could re-write <math>FG</math>. Indeed, we can: | ||
+ | <cmath>FG = AG - FG = \frac{r_a}{\tan \left( \frac{A}{2} \right)} - b \cos (A)</cmath> | ||
+ | and similarly for <math>FH</math>: | ||
+ | <cmath>FH = AH - AF = \frac{r_a}{\tan \left( \frac{A}{2} \right)} - c \cos (A)</cmath> | ||
+ | We now seek to re-write <math>FE</math>. Using the law of cosines: | ||
+ | <cmath>FE^2 = AF^2 + AE^ 2- 2AF \cdot AE \cos (A) = \left(c\cos (A) \right)^2 + \left(b\cos (A) \right)^2 - 2 (bc \cos ^2 (A)) \cos (A) = \cos ^2 (A) (c^2 + b^2 - 2bc \cos (A)) = \cos ^2 (A) \cdot a^2</cmath> | ||
+ | Therefore, | ||
+ | <cmath>FE = a \cos (A)</cmath> | ||
+ | Putting this all together, we see that: | ||
+ | <cmath>a \cos (A) = \frac{r_a}{\tan \left(\frac{a}{2} \right)} - b \cos (A) + \frac{r_a}{\tan \left(\frac{a}{2} \right)} - c \cos (A) \Longleftrightarrow</cmath> | ||
+ | <cmath>(a+b+c) \cos (A) = \frac{2r_a}{\tan \left(\frac{a}{2} \right)}</cmath> | ||
+ | Now, we seek to write <math>\tan \left(\frac{a}{2} \right)</math> in terms of the side lengths of the triangle. Notice that: | ||
+ | <cmath>2\cos ^2 \left(\frac{A}{2} \right)-1 = \cos (A) \Longleftrightarrow</cmath> | ||
+ | <cmath>2\cos ^2 \left(\frac{A}{2} \right) = 1 + \frac{b^2 + c^2-a^2}{2bc} = \frac{b^2 + 2bc + c^2 - a^2}{2bc} = \frac{(b+c)^2-a^2}{2bc} = \frac{(a+b+c)(b+c-a)}{2bc} = \frac{(2s)(2s-2a)}{2bc}</cmath> | ||
+ | where <math>s</math> is the semi-perimeter. We get that: | ||
+ | <cmath>\cos \left(\frac{A}{2} \right) = \sqrt{\frac{s(s-a)}{bc}}</cmath> | ||
+ | Using the fact that <math>\cos^2 \left(\frac{A}{2} \right) + \sin^2 \left(\frac{A}{2} \right) = 1</math>, we have that: | ||
+ | <cmath>\sin^2 \left(\frac{A}{2} \right) = 1-\frac{b^2 + 2bc + c^2 - a^2}{4bc} = \frac{(b-c)^2 - a^2}{4bc} = \frac{(2s-2c)(2s-2b)}{4bc}</cmath> | ||
+ | Therefore, | ||
+ | <cmath>\tan \left( \frac{A}{2} \right) = \sqrt{ \frac{(s-b)(s-c)}{s(s-a)} }</cmath> | ||
+ | Returning to our original problem: | ||
+ | <cmath>\frac{2r_a}{\sqrt{ \frac{(s-b)(s-c)}{s(s-a)} }} = (a+b+c) \cos (A)</cmath> | ||
+ | It is a well-known fact that <math>r_a = \sqrt{\frac{s(s-b)(s-c)}{s-a}}</math>, so: | ||
+ | <cmath>\frac{2\sqrt{\frac{s(s-b)(s-c)}{s-a}}}{\sqrt{ \frac{(s-b)(s-c)}{s(s-a)} }} = 2s \cos (A)</cmath> which implies that: | ||
+ | <cmath>\cos (A) =1 \Longleftrightarrow A = 0</cmath> | ||
+ | which is a contradiction. Hence, our original assumption that <math>EF</math> is tangent to the a-excircle is incorrect. <math>\blacksquare</math> | ||
+ | ~AopsUser101 | ||
==See also== | ==See also== | ||
+ | {{MAA Notice}} | ||
{{USAJMO newbox|year=2019|num-b=3|num-a=5}} | {{USAJMO newbox|year=2019|num-b=3|num-a=5}} |
Latest revision as of 20:27, 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 |