Difference between revisions of "2020 USOJMO Problems/Problem 4"
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==Solution 6== | ==Solution 6== | ||
− | Let <math>FE</math> meet <math>AC</math> at <math>J</math>, <math>BE</math> meet <math>AC</math> at <math>S</math>, connect <math>AE</math>. | + | Let <math>FE</math> meet <math>AC</math> at <math>J</math>, <math>BE</math> meet <math>AC</math> at <math>S</math>, connect <math>AE, SD</math>. |
Denote that <math>\angle{BCA}=\alpha; AB=BC, \angle{BAC}=\angle{BCA}=\alpha</math>, since <math>EF</math> is parallel to <math>BC</math>, <math>\angle{AJF}=\angle{ACB}=\alpha</math>. <math>\angle{AJF}</math>and <math>\angle{EJS}</math> are vertical angle, so they are equal to each other. | Denote that <math>\angle{BCA}=\alpha; AB=BC, \angle{BAC}=\angle{BCA}=\alpha</math>, since <math>EF</math> is parallel to <math>BC</math>, <math>\angle{AJF}=\angle{ACB}=\alpha</math>. <math>\angle{AJF}</math>and <math>\angle{EJS}</math> are vertical angle, so they are equal to each other. | ||
− | <math>BE\bot{AC}</math>,\angle{JES}=90^{\circ}-\alpha<math>, since < | + | <math>BE\bot{AC}</math>,<math>\angle{JES}=90^{\circ}-\alpha</math>, since <math>\angle{EFB}=\angle{AJF}+\angle{FAJ}=2\alpha</math>, we can express <math>\angle{FBE}=180^{\circ}-2\alpha-(90^{\circ}-\alpha)=90^{\circ}-\alpha= |
− | \ | + | \angle{FEB}</math>, leads to <math>FE=FB</math> |
+ | |||
+ | Notice that quadrilateral <math>AFED</math> is a cyclic quadrilateral since <math>\angle{ADE}+\angle{AFE}=\angle{ADE}+\angle{ABC}=180^{\circ}</math>. | ||
+ | |||
+ | Assume <math>\angle{ECA}=\beta</math>, <math>\triangle{AES}</math> is congruent to <math>\triangle{CES}</math> since <math>AS=AS,\angle{ASE}=\angle{BSE}, SE=SE(SAS)</math>, so we can get <math>\angle{EAS}=\beta</math> | ||
+ | Let the circumcircle of <math>AFED</math> meets <math>AC</math> at <math>Q</math> | ||
+ | Now notice that <math>\widehat{QE}=\widehat{QE}, \angle{QAE}=\angle{QDE}=\beta</math>; similarly, <math>\widehat{FQ}=\widehat{FQ}; \angle{FDQ}=\angle{FAQ}=\alpha</math>. | ||
+ | |||
+ | <math>\angle{FDE}=\alpha+\beta; \angle{FED}=\angle{BCD}=\alpha+\beta</math>, it leads to <math>FD=FE</math>. | ||
+ | |||
+ | since <math>FE=FB;FD=FE, DF=BF</math> as desired | ||
+ | ~bluesoul | ||
+ | |||
+ | ==See Also== | ||
+ | {{USAJMO newbox|year=2020|num-b=3|num-a=5}} | ||
+ | {{MAA Notice}} |
Latest revision as of 18:15, 6 October 2023
Contents
Problem
Let be a convex quadrilateral inscribed in a circle and satisfying . Points and are chosen on sides and such that and . Prove that .
Solution
Let be the intersection of and and be the intersection of and .
Claim:
By Pascal's on , we see that the intersection of and , , and are collinear. Since , we know that as well.
Note that since all cyclic trapezoids are isosceles, . Since and , we know that , from which we have that is an isosceles trapezoid and . It follows that , so is an isosceles trapezoid, from which , as desired.
Solution 2
Let , and let . Now let and .
From and , we have so . From cyclic quadrilateral ABCD, . Since , .
Now from cyclic quadrilateral ABC and we have . Thus F, A, D, and E are concyclic, and Let this be statement 1.
Now since , triangle ABC gives us . Thus , or .
Right triangle BHC gives , and implies
Now triangle BGE gives . But , so . Using triangle FGD and statement 1 gives
Thus, , so as desired.
~MortemEtInteritum
Solution 3 (Angle-Chasing)
Proving that is equivalent to proving that . Note that because quadrilateral is cyclic. Also note that because . , which follows from the facts that and , implies that . Thus, we would like to prove that triangle is similar to triangle . In order for this to be true, then must equal which implies that must equal . In order for this to be true, then quadrilateral must be cyclic. Using the fact that , we get that , and that , and thus we have proved that quadrilateral is cyclic. Therefore, triangle is similar to isosceles triangle from AA and thus .
-xXINs1c1veXx
Solution 4
BE is perpendicular bisector of AC, so . FE is parallel to BC and ABCD is cyclic, so AFED is also cyclic. . Hence, , .
Mathdummy
Solution 5
Let be on such that , and . Then is the orthic triangle of . Thus, is the midpoint of and lies on the bisector of .
Solution 6
Let meet at , meet at , connect . Denote that , since is parallel to , . and are vertical angle, so they are equal to each other. ,, since , we can express , leads to
Notice that quadrilateral is a cyclic quadrilateral since .
Assume , is congruent to since , so we can get Let the circumcircle of meets at Now notice that ; similarly, .
, it leads to .
since as desired ~bluesoul
See Also
2020 USAJMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.