Difference between revisions of "2018 USAJMO Problems/Problem 3"
Nukelauncher (talk | contribs) (Created page with "== Problem == (<math>*</math>) Let <math>ABCD</math> be a quadrilateral inscribed in circle <math>\omega</math> with <math>\overline{AC} \perp \overline{BD}</math>. Let <math>...") |
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==Solution 1== | ==Solution 1== | ||
+ | First we have that <math>BE=BD=BF</math> by the definition of a reflection. Let <math>\angle DEB = \alpha</math> and <math>\angle DFB = \beta.</math> Since <math>\triangle DBE</math> is isosceles we have <math>\angle BDE = \alpha.</math> Also, we see that <math>\angle BDE = \angle CAB = \angle CDB = \alpha,</math> using similar triangles and the property of cyclic quadrilaterals. Similarly, <cmath>\angle DFB = \angle FDB = \angle ACB = \angle ADB = \beta.</cmath> Now, from <math>BE=BD=BF</math> we know that <math>B</math> is the circumcenter of <math>\triangle DEF.</math> Using the properties of the circumcenter and some elementary angle chasing, we find that <cmath>\angle DPE = 90^{\circ} + \beta - \alpha.</cmath> | ||
+ | Now, we claim that <math>Q</math> is the intersection of ray <math>\overrightarrow{EB}</math> and the circumcircle of <math>ABCD.</math> To prove this, we just need to show that <math>DEPQ</math> is cyclic by this definition of <math>Q.</math> We have that <cmath>\angle DQE = \angle DCB = \angle DCA + \angle ACB = (90^{\circ}-\alpha)+\beta.</cmath> We also have from before that <cmath>\angle DPE = 90+\beta-\alpha,</cmath> so <math>\angle DQE=\angle DPE</math> and this proves the claim. | ||
+ | |||
+ | We can use a similar proof to show that <math>F, B, R</math> are collinear. | ||
+ | |||
+ | Now, <math>DP</math> is the radical axis of the circumcircles of <math>\triangle EDP</math> and <math>\triangle FDP.</math> Since <math>B</math> lies on <math>DP,</math> and <math>E, Q</math> lie on the circumcircle of <math>\triangle EPD</math> and <math>F, R</math> lie on the circumcircle of <math>\triangle FPD,</math> we have that <cmath>BF \cdot BR = BE \cdot BQ.</cmath> However, <math>BF=BE,</math> so <math>BR=BQ.</math> Since <math>E, B, Q</math> are collinear and so are <math>F, B, R</math> we can add these <math>2</math> equations to get <cmath>EQ=BE+BQ=BF+BR=FR,</cmath> which completes the proof. | ||
+ | |||
+ | ~nukelauncher | ||
+ | (Monday G. Fern) | ||
+ | |||
+ | ==Solution 2== | ||
+ | We begin with the following claims. | ||
+ | |||
+ | Claim. <math>B</math> is the circumcenter of <math>\triangle DEF</math>. | ||
+ | Proof. By reflection <math>BD=BE=BF</math>. | ||
+ | |||
+ | Claim. <math>A</math> is the circumcenter of <math>\triangle EPD</math>. | ||
+ | Proof. First, we have | ||
+ | <cmath>\angle DPE = 180^{\circ}-\angle PDE - \angle PED = 180^{\circ}-\angle BED - \frac{\angle DBF}{2} = \angle 180^{\circ}-\angle BED - \frac{180^{\circ}-2\angle BFD}{2} = 90^{\circ}-\angle BDE + \angle DFB</cmath> | ||
+ | |||
+ | Then | ||
+ | |||
+ | <cmath> \angle ADE = \angle ADP - \angle EDP = \angle ADB - \angle EDB = \angle ACB - \angle BDE = \angle FDB - \angle BDE = \angle DFB - \angle DEB = \angle DPE - 90^{\circ} = \frac{180^{\circ}-2(180^{\circ}-\angle DPE)}{2}</cmath> | ||
+ | Then <math>2\angle ADE = 180^{\circ}-2(180^{\circ}-\angle DPE) \implies \angle DAE = 2(180^{\circ}-\angle DPE)</math>. This is enough to imply what we desire. \newline | ||
+ | |||
+ | Claim. <math>C</math> is the circumcenter of <math>\triangle FPD</math>. | ||
+ | Proof. Similar to above. | ||
+ | |||
+ | Claim. <math>E, B, Q</math> are collinear. | ||
+ | Proof. We have | ||
+ | <cmath>\angle ABE = \angle ABD = \angle AQD = \angle ADQ = 180^{\circ}-\angle ABQ</cmath> | ||
+ | |||
+ | Claim.<math>F, B, R</math> are collinear. | ||
+ | Proof. Similar to above. | ||
+ | |||
+ | Since <math>DEPQ</math> is cyclic, <math>\triangle EBP \sim \triangle DBQ</math>. However, <math>BE=BD</math>, so <math>BP=BQ</math>. Similarly, <math>BP=BR</math>. Finishing, we have <cmath>EQ = EB + BQ = BD + BP = FB + BR = FR</cmath>, as desired. <math>\blacksquare</math> ~MSC | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 19:30, 4 November 2020
Contents
Problem
() Let be a quadrilateral inscribed in circle with . Let and be the reflections of over lines and , respectively, and let be the intersection of lines and . Suppose that the circumcircle of meets at and , and the circumcircle of meets at and . Show that .
Solution 1
First we have that by the definition of a reflection. Let and Since is isosceles we have Also, we see that using similar triangles and the property of cyclic quadrilaterals. Similarly, Now, from we know that is the circumcenter of Using the properties of the circumcenter and some elementary angle chasing, we find that
Now, we claim that is the intersection of ray and the circumcircle of To prove this, we just need to show that is cyclic by this definition of We have that We also have from before that so and this proves the claim.
We can use a similar proof to show that are collinear.
Now, is the radical axis of the circumcircles of and Since lies on and lie on the circumcircle of and lie on the circumcircle of we have that However, so Since are collinear and so are we can add these equations to get which completes the proof.
~nukelauncher (Monday G. Fern)
Solution 2
We begin with the following claims.
Claim. is the circumcenter of . Proof. By reflection .
Claim. is the circumcenter of . Proof. First, we have
Then
Then . This is enough to imply what we desire. \newline
Claim. is the circumcenter of . Proof. Similar to above.
Claim. are collinear. Proof. We have
Claim. are collinear. Proof. Similar to above.
Since is cyclic, . However, , so . Similarly, . Finishing, we have , as desired. ~MSC
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
See also
2018 USAJMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |