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[[Category:MA460_(Fall2009Walther)]] | [[Category:MA460_(Fall2009Walther)]] | ||
| − | On number 5, I have been able to prove the => for all three of them, but I am struggling with the <= for (i) and (ii). Any hints on where to start?-Lauren | + | * Number four seems to be a nightmare. Menelaus can be applied at least five different ways. And it takes forever to work out which points correspond to the points in the formula. Anyone have more specific information on this problem, such as which triangles actually yield the required ratios? I have everything except 3 and 4. |
| + | ** For any unfortunate souls up this late still doing this horrible problem, I imagine you have probably undergone frustration sufficient to have earned a solution. The two triangles to which Menelaus is applied are CC'B and ACC'. | ||
| + | -From this I went about and created two equations with Menelaus formula and then i set the two equations to be equal to each other which gave me the end result of what we were trying to prove. -Brian | ||
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| + | Does anybody have anything for number 3?- Mark | ||
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| + | On number 5, I have been able to prove the => for all three of them, but I am struggling with the <= for (i) and (ii). Any hints on where to start?-Lauren | ||
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| + | Prove that the angle bisectors are unequal by showing that the triangles they create can't be congruent. | ||
Lauren, for (i)<= use the fact that altitudes are perpendicular so we have right triangles. They have an equal leg (the base of the big triangle) and the hypotenuse of both are equal (given-they are the two altitudes). Now use Theorem 9 to prove conguent triangles and then the base angles of the big triangle are equal-then use theorem 5. Hope this helps. ~Janelle | Lauren, for (i)<= use the fact that altitudes are perpendicular so we have right triangles. They have an equal leg (the base of the big triangle) and the hypotenuse of both are equal (given-they are the two altitudes). Now use Theorem 9 to prove conguent triangles and then the base angles of the big triangle are equal-then use theorem 5. Hope this helps. ~Janelle | ||
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for #5, 3iii <=, DE and AB are parallel since they are medians. So big triangle and inner triangle are similar. so 2 bottom angles are equal so is an iso triangle. - Sue | for #5, 3iii <=, DE and AB are parallel since they are medians. So big triangle and inner triangle are similar. so 2 bottom angles are equal so is an iso triangle. - Sue | ||
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| + | I don't think we know that the 2 "bottom" angles are equal in the inner triangle. I am not sure how you proved that.--[[User:Jrhaynie|Jrhaynie]] 16:31, 30 September 2009 (UTC) | ||
for ii <=, can we do theorem 21? we know some equal angles. This would make for some equal sides if the sines are equal and than work around that? - Sue | for ii <=, can we do theorem 21? we know some equal angles. This would make for some equal sides if the sines are equal and than work around that? - Sue | ||
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For #4, try looking for different instances of Menalaus' theorem inside the triangle and combine them. -Tim | For #4, try looking for different instances of Menalaus' theorem inside the triangle and combine them. -Tim | ||
| + | * I cannot help but wonder at this suggestion. It is like being told that the needle is in the third haystack from the left. | ||
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| + | Does anyone have a hint for 3? - Dana | ||
For 7, use Thm 16 to makes two different sets of lines parallel. -Mary | For 7, use Thm 16 to makes two different sets of lines parallel. -Mary | ||
| + | I am having trouble with 6, is it similar to case 1? --[[User:Jrhaynie|Jrhaynie]] 16:15, 30 September 2009 (UTC) | ||
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| + | #6 can be solved relatively similar to case 1, you just have to use different similar triangles. -Kat | ||
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| + | Does anyone have a hint for number 3? I started with a parallelogram with sides congruent to the diagonals of ABCD, but not sure where to go next! | ||
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| + | So if you connect the diagonals you can find the | ||
| + | areas of the small triangles relative to the area | ||
| + | of half of ABCD and you do this for all 4 small | ||
| + | triangles and the remaining area must equal MNPQ. | ||
| + | --[[User:Jrhaynie|Jrhaynie]] 08:29, 1 October 2009 (UTC) | ||
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Latest revision as of 08:33, 1 October 2009
- Number four seems to be a nightmare. Menelaus can be applied at least five different ways. And it takes forever to work out which points correspond to the points in the formula. Anyone have more specific information on this problem, such as which triangles actually yield the required ratios? I have everything except 3 and 4.
- For any unfortunate souls up this late still doing this horrible problem, I imagine you have probably undergone frustration sufficient to have earned a solution. The two triangles to which Menelaus is applied are CC'B and ACC'.
-From this I went about and created two equations with Menelaus formula and then i set the two equations to be equal to each other which gave me the end result of what we were trying to prove. -Brian
Does anybody have anything for number 3?- Mark
On number 5, I have been able to prove the => for all three of them, but I am struggling with the <= for (i) and (ii). Any hints on where to start?-Lauren
Prove that the angle bisectors are unequal by showing that the triangles they create can't be congruent.
Lauren, for (i)<= use the fact that altitudes are perpendicular so we have right triangles. They have an equal leg (the base of the big triangle) and the hypotenuse of both are equal (given-they are the two altitudes). Now use Theorem 9 to prove conguent triangles and then the base angles of the big triangle are equal-then use theorem 5. Hope this helps. ~Janelle
Lauren, I am having trouble with the <= for (iii) could you help me out there? ~Janelle
for #5, 3iii <=, DE and AB are parallel since they are medians. So big triangle and inner triangle are similar. so 2 bottom angles are equal so is an iso triangle. - Sue
I don't think we know that the 2 "bottom" angles are equal in the inner triangle. I am not sure how you proved that.--Jrhaynie 16:31, 30 September 2009 (UTC)
for ii <=, can we do theorem 21? we know some equal angles. This would make for some equal sides if the sines are equal and than work around that? - Sue
Sue, since we only have to do two of the three, you might want to do i and iii. These proofs are much easier. -Jennie
Does anyone have hints for #3,4, or 7?
For #4, try looking for different instances of Menalaus' theorem inside the triangle and combine them. -Tim
- I cannot help but wonder at this suggestion. It is like being told that the needle is in the third haystack from the left.
Does anyone have a hint for 3? - Dana
For 7, use Thm 16 to makes two different sets of lines parallel. -Mary
I am having trouble with 6, is it similar to case 1? --Jrhaynie 16:15, 30 September 2009 (UTC)
- 6 can be solved relatively similar to case 1, you just have to use different similar triangles. -Kat
Does anyone have a hint for number 3? I started with a parallelogram with sides congruent to the diagonals of ABCD, but not sure where to go next!
So if you connect the diagonals you can find the
areas of the small triangles relative to the area
of half of ABCD and you do this for all 4 small
triangles and the remaining area must equal MNPQ.
--Jrhaynie 08:29, 1 October 2009 (UTC)
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