Difference between revisions of "2018 USAJMO Problems/Problem 3"
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+ | (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 20:30, 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 |