Difference between revisions of "2020 USOJMO Problems/Problem 4"
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+ | ==Problem== | ||
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+ | Let <math>ABCD</math> be a convex quadrilateral inscribed in a circle and satisfying <math>DA < AB = BC < CD</math>. Points <math>E</math> and <math>F</math> are chosen on sides <math>CD</math> and <math>AB</math> such that <math>BE \perp AC</math> and <math>EF \parallel BC</math>. Prove that <math>FB = FD</math>. | ||
+ | |||
+ | ==Solution== | ||
Let <math>G</math> be the intersection of <math>AE</math> and <math>(ABCD)</math> and <math>H</math> be the intersection of <math>DF</math> and <math>(ABCD)</math>. | Let <math>G</math> be the intersection of <math>AE</math> and <math>(ABCD)</math> and <math>H</math> be the intersection of <math>DF</math> and <math>(ABCD)</math>. | ||
− | + | ||
+ | <b>Claim: <math>GH || FE || BC</math></b> | ||
+ | |||
By Pascal's on <math>GDCBAH</math>, we see that the intersection of <math>GH</math> and <math>BC</math>, <math>E</math>, and <math>F</math> are collinear. Since <math>FE || BC</math>, we know that <math>HG || BC</math> as well. <math>\blacksquare</math> | By Pascal's on <math>GDCBAH</math>, we see that the intersection of <math>GH</math> and <math>BC</math>, <math>E</math>, and <math>F</math> are collinear. Since <math>FE || BC</math>, we know that <math>HG || BC</math> as well. <math>\blacksquare</math> | ||
− | + | ||
Note that since all cyclic trapezoids are isosceles, <math>HB = GC</math>. Since <math>AB = BC</math> and <math>EB \perp AC</math>, we know that <math>EA = EC</math>, from which we have that <math>DGCA</math> is an isosceles trapezoid and <math>DA = GC</math>. It follows that <math>DA = GC = HB</math>, so <math>BHAD</math> is an isosceles trapezoid, from which <math>FB = FD</math>, as desired. <math>\blacksquare</math> | Note that since all cyclic trapezoids are isosceles, <math>HB = GC</math>. Since <math>AB = BC</math> and <math>EB \perp AC</math>, we know that <math>EA = EC</math>, from which we have that <math>DGCA</math> is an isosceles trapezoid and <math>DA = GC</math>. It follows that <math>DA = GC = HB</math>, so <math>BHAD</math> is an isosceles trapezoid, from which <math>FB = FD</math>, as desired. <math>\blacksquare</math> | ||
+ | |||
+ | ==Solution 2== | ||
+ | Let <math>G=\overline{FE}\cap\overline{BC}</math>, and let <math>G=\overline{AC}\cap\overline{BE}</math>. Now let <math>x=\angle ACE</math> and <math>y=\angle BCA</math>. | ||
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+ | From <math>BA=BC</math> and <math>\overline{BE}\perp \overline{AC}</math>, we have <math>AE=EC</math> so <math>\angle EAC =\angle ECA = x</math>. From cyclic quadrilateral ABCD, <math>\angle ABD = \angle ACD = x</math>. Since <math>BA=BC</math>, <math>\angle BCA = \angle BAC = y</math>. | ||
+ | |||
+ | Now from cyclic quadrilateral ABC and <math>\overline{FE}\parallel \overline{BC}</math> we have <math>\angle FAC = \angle BAC = \pi - \angle BCD = \pi - \angle FED</math>. Thus F, A, D, and E are concyclic, and <math>\angle DFG = \angle DAE = \angle DAC - \angle EAC = \angle DBC - x</math> Let this be statement 1. | ||
+ | |||
+ | Now since <math>\overline{AH}\perp \overline {BH}</math>, triangle ABC gives us <math>\angle BAH + \angle ABG = \frac{\pi}{2}</math>. Thus <math>y+x+\angle GBE=\frac{\pi}{2}</math>, or <math>\angle GBE = \frac{\pi}{2}-x-y</math>. | ||
+ | |||
+ | Right triangle BHC gives <math>\angle HBC = \frac{\pi}{2}-y</math>, and <math>\overline{BC}\parallel \overline{FE}</math> implies <math>\angle BEG=\angle HBC = \frac{\pi}{2}-y.</math> | ||
+ | |||
+ | Now triangle BGE gives <math>\angle BGE = \pi - \angle BEG - \angle GBE = \pi - (\frac{\pi}{2}-y)-(\frac{\pi}{2}-x-y)=x+2y</math>. But <math>\angle FGB = \angle BGE</math>, so <math>\angle FGB=x+2y</math>. Using triangle FGD and statement 1 gives <cmath>\begin{align*}\angle FDG &= \pi - \angle DFG - \angle FGB \\ | ||
+ | &= \pi - (\angle DBC - x) - (x + 2y) \\ | ||
+ | &= \pi - (\angle GBE + \angle EBC - x) - (x + 2y) \\ | ||
+ | &= \pi - ([\frac{\pi}{2}-x-y]+[\frac{\pi}{2}-y]-x)-(x+2y) \\ | ||
+ | &= x \\ | ||
+ | &= \angle FBD\end{align*}</cmath> | ||
+ | |||
+ | Thus, <math>\angle FDB = \angle FBD</math>, so <math>\boxed{FB=FD}</math> as desired.<math>\blacksquare</math> | ||
+ | |||
+ | ~MortemEtInteritum | ||
+ | |||
+ | ==Solution 3 (Angle-Chasing)== | ||
+ | Proving that <math>FB=FD</math> is equivalent to proving that <math>\angle FBD= \angle FDB</math>. Note that <math>\angle FBD=\angle ACD</math> because quadrilateral <math>ABCD</math> is cyclic. Also note that <math>\angle BAC=\angle ACB</math> because <math>AB=BC</math>. <math>AE=EC</math>, which follows from the facts that <math>BE \perp AC</math> and <math>AB=AC</math>, implies that <math>\angle CAE= \angle ACE= \angle ACD= \angle FBD</math>. Thus, we would like to prove that triangle <math>FBD</math> is similar to triangle <math>AEC</math>. In order for this to be true, then <math>\angle BFD</math> must equal <math>\angle AEC</math> which implies that <math>\angle AFD</math> must equal <math>\angle AED</math>. In order for this to be true, then quadrilateral <math>AFED</math> must be cyclic. Using the fact that <math>EF \parallel BC</math>, we get that <math>\angle AFE= \angle ABC</math>, and that <math>\angle FED= \angle BCE</math>, and thus we have proved that quadrilateral <math>AFED</math> is cyclic. Therefore, triangle <math>FDB</math> is similar to isosceles triangle <math>AEC</math> from AA and thus <math>FB=FD</math>. | ||
+ | |||
+ | -xXINs1c1veXx | ||
+ | ==Solution 4 == | ||
+ | BE is perpendicular bisector of AC, so <math>\angle ACE = \angle EAC</math>. FE is parallel to BC and ABCD is cyclic, so AFED is also cyclic. <math>\angle AFD = \angle AED = 2 \angle ACE = 2 \angle FBD</math>. Hence, <math>\angle FBD = \angle BDF</math>, <math> FD = FB</math>. | ||
+ | |||
+ | Mathdummy |
Revision as of 02:13, 29 October 2020
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