Difference between revisions of "2007 JBMO Problems/Problem 2"
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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> | ||
==See Also== | ==See Also== | ||
| − | {{JBMO box|year=2007|before=1|num-a=3|five=}} | + | {{JBMO box|year=2007|before-a=1|num-a=3|five=}} |
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
Revision as of 08:13, 2 April 2024
Problem 2
Let 
be a convex quadrilateral with 
, 
and 
. The diagonals and intersect at point 
. Determine the measure of 
.
Solution
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 
See Also
| 2007 JBMO (Problems • Resources) | ||
| Preceded by [[2007 JBMO Problems/Problem {{{num-b}}}|Problem {{{num-b}}}]] |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 | ||
| All JBMO Problems and Solutions | ||
