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= | + | {{JBMO box|year=2007|before=1|num-a=3|five=}} |
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] |
Revision as of 09:13, 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 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 | ||
All JBMO Problems and Solutions |