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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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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> i steh south pole of <math>A</math> so <math>I</math> is on the circle of <math>DAC</math> but we can easily seen that not possible
+
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 09:29, 10 January 2024

Let I be the intersection between $(DP)$ and the angle bisector of $\angle{DAP}$ So $\angle{CAI}=\angle{PAI}=36/2°=18°$ So $\angle{CAI}=18°=\angle{CBD}=\angle{CBI}$ We can conclude that $A,B,C,I$ are on a same circle. So $\angle{ICB}=180-\angle{IAB}=180-\angle{IAC}-\angle{CAB}=180-18-72=90$ Because $\angle{CBD}=18$ and $\angle{CDB}=36$ we have $126=\angle{DCB}=\angle{ICB}+\angle{ICD}=90+\angle{ICD}$ So $\angle{ICD}=36=\angle{BDC}=\angle{IDC}$ So $I$ is on the angle bisector of $\angle{DAP}$ and on the mediator of $DC$. The first posibility is that $I$ is the south pole of $A$ so $I$ is on the circle of $DAC$ but we can easily seen that's not possible The second possibility is that $DAC$ is isosceles in $A$. So because $\angle{DAC}=36$ and $DAC$ is isosceles in $A$ we have $\angle{ADC}=72$. So $\angle{APD}=180-\angle{PAD}-\angle{PDA}=180-36-(\angle{ADC}-\angle{PDC})=180-36-72+36=108$

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