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 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 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 | ||
All JBMO Problems and Solutions |