Difference between revisions of "Talk:2009 AIME I Problems/Problem 15"
(New page: Sorry, I don't know LaTeX but I do think I still have this problem's solution, which hasn't yet been posted. Maybe someone can validate / compare, and format things nicely afterward. Lem...) |
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| − | Sorry, I don't know LaTeX but I | + | Sorry, I don't know LaTeX, but I saw that there was no solution/answer for this problem yet. Maybe someone can validate / compare to my own, and format things nicely afterward. |
Lemma: Angle BPC = 150deg regardless of choice of D | Lemma: Angle BPC = 150deg regardless of choice of D | ||
Proof: | Proof: | ||
| − | + | Call angles: | |
| + | IbBD = ABIb = a1, | ||
BDIb = IbDA = a2, | BDIb = IbDA = a2, | ||
IcDC = ADIc = a3, | IcDC = ADIc = a3, | ||
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1) We can then write that angle BAC = (180-2a1-2a2) + (180-2a3-2a4) = 360-2(a1+a2+a3+a4) | 1) We can then write that angle BAC = (180-2a1-2a2) + (180-2a3-2a4) = 360-2(a1+a2+a3+a4) | ||
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2) angle BAC = 60deg using law of cosines on the original triangle | 2) angle BAC = 60deg using law of cosines on the original triangle | ||
| − | 1) and 2) mean that (a1+a2+a3+a4) = 150deg | + | *-> 1) and 2) mean that (a1+a2+a3+a4) = 150deg |
| − | + | Quads BPDIb and DPCIc are both cyclical, by problem definition. The property of cyclical quads that will be used is the fact that opposite angles are supplementary. | |
3) angle BPD = 180 - angle BIbD | 3) angle BPD = 180 - angle BIbD | ||
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4) angle BIbD = 180-a1-a2 | 4) angle BIbD = 180-a1-a2 | ||
| − | 3) and 4) mean that angle BPD = a1+a2 | + | *-> 3) and 4) mean that angle BPD = a1+a2 |
Similarly, angle DPC = a3+a4 | Similarly, angle DPC = a3+a4 | ||
Latest revision as of 18:52, 25 March 2009
Sorry, I don't know LaTeX, but I saw that there was no solution/answer for this problem yet. Maybe someone can validate / compare to my own, and format things nicely afterward.
Lemma: Angle BPC = 150deg regardless of choice of D
Proof: Call angles:
IbBD = ABIb = a1,
BDIb = IbDA = a2,
IcDC = ADIc = a3,
DCIc = IcCA = a4.
(the pairs of angles are equal due to Ib, Ic being the incenters, implying that segments BIb and CIc are angle bisectors of angles ABD and DCA, respectively)
1) We can then write that angle BAC = (180-2a1-2a2) + (180-2a3-2a4) = 360-2(a1+a2+a3+a4)
2) angle BAC = 60deg using law of cosines on the original triangle
- -> 1) and 2) mean that (a1+a2+a3+a4) = 150deg
Quads BPDIb and DPCIc are both cyclical, by problem definition. The property of cyclical quads that will be used is the fact that opposite angles are supplementary.
3) angle BPD = 180 - angle BIbD
4) angle BIbD = 180-a1-a2
- -> 3) and 4) mean that angle BPD = a1+a2
Similarly, angle DPC = a3+a4
Finally, angle BPC = angle BPD + angle DPC = a1+a2+a3+a4 = 150deg
The problem then reduces to finding the maximum area of a triangle with base 14 and opposite angle 150deg, which can be proven any number of ways to be when the triangle is isosceles.
This gives an area of 98-49sqrt(3), and a problem answer of (98+49+3) = 150
