Difference between revisions of "2017 USAJMO Problems/Problem 5"
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==Solution 1== | ==Solution 1== | ||
It's well known that the reflection of <math>H</math> across <math>\overline{BC}</math>, <math>H'</math>, lies on <math>(ABC)</math>. Then <math>(BHC)</math> is just the reflection of <math>(BH'C)</math> across <math>\overline{BC}</math>, which is equivalent to the reflection of <math>(ABC)</math> across <math>\overline{BC}</math>. Reflect points <math>A</math> and <math>N</math> across <math>\overline{BC}</math> to points <math>A'</math> and <math>N'</math>, respectively. Then <math>N'</math> is the midpoint of minor arc <math>\overarc{BC}</math>, so <math>A, D, N'</math> are collinear in that order. It suffices to show that <math>\angle AA'N'=\angle ADO</math>. | It's well known that the reflection of <math>H</math> across <math>\overline{BC}</math>, <math>H'</math>, lies on <math>(ABC)</math>. Then <math>(BHC)</math> is just the reflection of <math>(BH'C)</math> across <math>\overline{BC}</math>, which is equivalent to the reflection of <math>(ABC)</math> across <math>\overline{BC}</math>. Reflect points <math>A</math> and <math>N</math> across <math>\overline{BC}</math> to points <math>A'</math> and <math>N'</math>, respectively. Then <math>N'</math> is the midpoint of minor arc <math>\overarc{BC}</math>, so <math>A, D, N'</math> are collinear in that order. It suffices to show that <math>\angle AA'N'=\angle ADO</math>. | ||
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<b>Claim:</b> <math>\triangle AA'N' \sim \triangle ADO</math>. The proof easily follows. | <b>Claim:</b> <math>\triangle AA'N' \sim \triangle ADO</math>. The proof easily follows. | ||
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<b>Proof:</b> Note that <math>\angle BAA'=\angle CAO=90^{\circ}-\angle ABC</math>. Then we have <math>\angle A'AN=\angle BAD-\angle BAA'=\angle CAD-\angle CAO=\angle DAO</math>. So, it suffices to show that <cmath>\frac{AA'}{AN'}=\frac{AD}{AO}\rightarrow AA'\cdot AO=AN'\cdot AD.</cmath> Notice that <math>\triangle ABA' \sim \triangle AOC</math>, so that <cmath>\frac{AB}{AA'}=\frac{AO}{AC}\rightarrow AA'\cdot AO=AB\cdot AC.</cmath> Therefore, it suffices to show that <cmath>AB\cdot AC=AN'\cdot AD\rightarrow \frac{AB}{AN'}=\frac{AD}{AC}.</cmath> But it is easy to show that <math>\triangle BAN\sim \triangle DAC</math>, implying the result. <math>\blacksquare</math> | <b>Proof:</b> Note that <math>\angle BAA'=\angle CAO=90^{\circ}-\angle ABC</math>. Then we have <math>\angle A'AN=\angle BAD-\angle BAA'=\angle CAD-\angle CAO=\angle DAO</math>. So, it suffices to show that <cmath>\frac{AA'}{AN'}=\frac{AD}{AO}\rightarrow AA'\cdot AO=AN'\cdot AD.</cmath> Notice that <math>\triangle ABA' \sim \triangle AOC</math>, so that <cmath>\frac{AB}{AA'}=\frac{AO}{AC}\rightarrow AA'\cdot AO=AB\cdot AC.</cmath> Therefore, it suffices to show that <cmath>AB\cdot AC=AN'\cdot AD\rightarrow \frac{AB}{AN'}=\frac{AD}{AC}.</cmath> But it is easy to show that <math>\triangle BAN\sim \triangle DAC</math>, implying the result. <math>\blacksquare</math> | ||
Revision as of 17:29, 21 September 2021
Contents
[hide]Problem
Let and be the circumcenter and the orthocenter of an acute triangle . Points and lie on side such that and . Ray intersects the circumcircle of triangle in point . Prove that .
Solution 1
It's well known that the reflection of across , , lies on . Then is just the reflection of across , which is equivalent to the reflection of across . Reflect points and across to points and , respectively. Then is the midpoint of minor arc , so are collinear in that order. It suffices to show that .
Claim: . The proof easily follows.
Proof: Note that . Then we have . So, it suffices to show that Notice that , so that Therefore, it suffices to show that But it is easy to show that , implying the result.
Solution 2
Suppose ray intersects the circumcircle of at , and let the foot of the A-altitude of be . Note that . Likewise, . So, . is cyclic, so . Also, . These two angles are on different circles and have the same measure, but they point to the same line ! Hence, the two circles must be congruent. (This is also a well-known result)
We know, since is the midpoint of , that is perpendicular to . is also perpendicular to , so the two lines are parallel. is a transversal, so . We wish to prove that , which is equivalent to being cyclic.
Now, assume that ray intersects the circumcircle of at a point . Point must be the midpoint of . Also, since is an angle bisector, it must also hit the circle at the point . The two circles are congruent, which implies NDP is isosceles. Angle ADN is an exterior angle, so . Assume WLOG that . So, . In addition, . Combining these two equations, .
Opposite angles sum to , so quadrilateral is cyclic, and the condition is proved.
-william122
See also
2017 USAJMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |