Difference between revisions of "2020 USOJMO Problems/Problem 4"
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==Solution 5== | ==Solution 5== | ||
Let <math>G</math> be on <math>AB</math> such that <math>GD \perp DB</math>, and <math>H = GD \cap EB</math>. Then <math>\angle{ADB} = \angle{EDB} = 90^{\circ} - \angle{ABE} \implies \triangle{DAE}</math> is the orthic triangle of <math>\triangle{HGB}</math>. Thus, <math>F</math> is the midpoint of <math>GB</math> and lies on the <math>\perp</math> bisector of <math>DB</math>. | Let <math>G</math> be on <math>AB</math> such that <math>GD \perp DB</math>, and <math>H = GD \cap EB</math>. Then <math>\angle{ADB} = \angle{EDB} = 90^{\circ} - \angle{ABE} \implies \triangle{DAE}</math> is the orthic triangle of <math>\triangle{HGB}</math>. Thus, <math>F</math> is the midpoint of <math>GB</math> and lies on the <math>\perp</math> bisector of <math>DB</math>. | ||
+ | |||
+ | ==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>. | ||
+ | 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>\angle{EFB}=\angle{AJF}+\angle{FAJ}=2\alpha<math>, we can express </math>\angel{FBE}=180^{\circ}-2\alpha-(90^{\circ}-\alpha)=90^{\circ}-\alpha= | ||
+ | \angel{FEB}--FE=FB$ |
Revision as of 04:21, 9 January 2022
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.
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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 .
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Solution 4
BE is perpendicular bisector of AC, so . FE is parallel to BC and ABCD is cyclic, so AFED is also cyclic. . Hence, , .
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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. ,\angle{JES}=90^{\circ}-\alpha\angle{EFB}=\angle{AJF}+\angle{FAJ}=2\alpha\angel{FBE}=180^{\circ}-2\alpha-(90^{\circ}-\alpha)=90^{\circ}-\alpha= \angel{FEB}--FE=FB$