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Difference between revisions of "2007 JBMO Problems/Problem 2"
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==Problem 2==
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Let <math>ABCD</math> be a convex quadrilateral with <math>\angle{DAC}= \angle{BDC}= 36^\circ</math> , <math>\angle{CBD}= 18^\circ</math> and <math>\angle{BAC}= 72^\circ</math>. The diagonals and intersect at point <math>P</math> . Determine the measure of <math>\angle{APD}</math>.
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==Solution==
 
Let I be the intersection between <math>(DP)</math> and the angle bisector of <math>\angle{DAP}</math>  
 
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}=\angle{PAI}=36/2°=18°</math>
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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 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>
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==See Also==
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{{JBMO box|year=2007|num-b=1|num-a=3|five=}}
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[[Category:Intermediate Geometry Problems]]

Latest revision as of 08:14, 2 April 2024

Problem 2

Let $ABCD$ be a convex quadrilateral with $\angle{DAC}= \angle{BDC}= 36^\circ$ , $\angle{CBD}= 18^\circ$ and $\angle{BAC}= 72^\circ$. The diagonals and intersect at point $P$ . Determine the measure of $\angle{APD}$.

Solution

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 $\angle{APD}=180-\angle{PAD}-\angle{PDA}=180-36-(\angle{ADC}-\angle{PDC})=180-36-72+36=108$

See Also

2007 JBMO (Problems • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4
All JBMO Problems and Solutions
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