Difference between revisions of "2007 JBMO Problems/Problem 2"
(Created page with "Let I be the intersection between <math>(DP)</math> and the angle bisector of <math>\angle{DAP}</math> So <math>\angle{CAI}=\angle{PAI}=36/2°=18°</math> So <math>\angle{CAI...") |
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Line 7: | Line 7: | ||
So <math>\angle{ICD}=36=\angle{BDC}=\angle{IDC}</math> | So <math>\angle{ICD}=36=\angle{BDC}=\angle{IDC}</math> | ||
So <math>I</math> is on the angle bisector of <math>\angle{DAP}</math> and on the mediator of <math>DC</math>. | So <math>I</math> is on the angle bisector of <math>\angle{DAP}</math> and on the mediator of <math>DC</math>. | ||
− | The first posibility is that <math>I</math> | + | The first posibility is that <math>I</math> is the south pole of <math>A</math> so <math>I</math> is on the circle of <math>DAC</math> but we can easily seen that's not possible |
The second possibility is that <math>DAC</math> is isosceles in <math>A</math>. So because <math>\angle{DAC}=36</math> and <math>DAC</math> is isosceles in <math>A</math> we have <math>\angle{ADC}=72</math>. So <math>\angle{APD}=180-\angle{PAD}-\angle{PDA}=180-36-(\angle{ADC}-\angle{PDC})=180-36-72+36=108</math> | The second possibility is that <math>DAC</math> is isosceles in <math>A</math>. So because <math>\angle{DAC}=36</math> and <math>DAC</math> is isosceles in <math>A</math> we have <math>\angle{ADC}=72</math>. So <math>\angle{APD}=180-\angle{PAD}-\angle{PDA}=180-36-(\angle{ADC}-\angle{PDC})=180-36-72+36=108</math> |
Revision as of 10:29, 10 January 2024
Let I be the intersection between and the angle bisector of
So
So
We can conclude that
are on a same circle.
So
Because
and
we have
So
So
is on the angle bisector of
and on the mediator of
.
The first posibility is that
is the south pole of
so
is on the circle of
but we can easily seen that's not possible
The second possibility is that
is isosceles in
. So because
and
is isosceles in
we have
. So