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<p>I love how he writes! It’s very hot! lol</p>
<p>And I’m neither gay nor female if you’re wondering. lol</p>
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<p>Don’t you mean his SAT score is hot?</p>
<p>lol. nah, jk.</p>
<p>Anyways, here’s a question. Bet you can’t solve this under a minute. :)</p>
<p><a href=“http://i56.■■■■■■■■■■■■■■■/albums/g178/xokuroxo92/mathquestionhard.png[/url]”>http://i56.■■■■■■■■■■■■■■■/albums/g178/xokuroxo92/mathquestionhard.png</a></p>
![http://i56.■■■■■■■■■■■■■■■/albums/g178/xokuroxo92/mathquestionhard.png[/url]](http://i56.photobucket.com/albums/g178/xokuroxo92/mathquestionhard.png%5B/url%5D){kind=link}
<p>If the figure is accurate, the answer’s POSSIBLY (e) 1:12. And that took 3 minutes since I was trying to approach it formulaically. Didn’t work. :(</p>
<p>^Correct~ it would be e and the diagram is correct also. :)</p>
<p>OK this took me 10 min, and then another 10 to reorganize everything and 15 to type it up in order. Call me slow.</p>
<p>What I did was lable PN and QN as X, QN as 2x, PM and RM as Y and PR as 2Y. </p>
<p>Triangle NRP and NRQ share NR, and they both have X as the other leg, so QR=2Y. QMP and QMR share QM, both have Y as the other leg, so 2X=2Y.</p>
<p>I replaced all the Y’s with X’s, and used “line connecting midpoints of 2 sides is paralell to 3rd side and is half of the 3rd” to make MN = X. Triangles MNP and QRP are both equilateral.</p>
<p>since Triangles NMS and QSR are similar Triangles, NS= 1/2 SR and MS = 1/2 QM. Then, I noticed that Triangle QNR has QN= X and QR = 2x, QNR is a 30-60-90 triangle. From that, I got <SNM = <SMN = <SQR= <SRQ = 30, so <NSM = 120 and so does <QSR via vert <'s. NR, opposite 60, must be Xsqrt3. Using the simlar triangles, NS = MS = X/sqrt3, so SR = QS = 2x/sqrt3. Height of Triangle QSR must = half of QS because 30-60-90, so Height of Triangle QSR= x/sqrt3. Using similar Triangles, the H of triangle MNS = X/2sqrt3.</p>
<p>Thus, AofMNS = X^2 / 2sqrt3. </p>
<p>Height of triangle QPR(an equilateral triangle) was obvious from the beginning; it’s Xsqrt3. So A of Triangle QPR = (2sqrt3)(X^2)</p>
<p>X^2 / 2sqrt3 divided by (2sqrt3)(X^2), the X^2’s cancel out, leaving 1 over (2sqrt3)^2, which is 1 over (4 times 3), = 1/12 E</p>
<p>Does any of that match up with your answer key ruvuitton?</p>
<p>I used side ratios to get NM/QR = 1/2</p>
<p>From the above I found the respective height ratios: 1:2 for NMS to QSR. </p>
<p>I joined P to the midpoint of QR. This lines passes through the centroid, which allows us to deduce (through side ratios) that the height ratio for triangle NMS to triangle PQR is 1:6.</p>
<p>Using the Area formula yielded: Triangle NMS: 1/2 * 1/2 * 1/6 = 1/24 units
Triangle PQR: 1/2 * 1 * 1 = 1/2 units
Therefore, the ratio of areas: 1:12</p>
<p>Edit: Jumbled expression a bit</p>
<p>You’ve gotta be kidding me.</p>
<p>Yea, 10 min to solve an SAT problem when I could’ve just learned centroids.</p>
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<p>I don’t think you can prove QR=2Y with this method. 2 sides are insufficient for a conqruency proof.</p>
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<p>Can you explain what you did here?</p>
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<p>Wait I thought that if the two sides make up the Hyp and a Leg, then the congruency can be proved? Havn’t looked @ geometry in a few years.</p>
<p>For that proof you would also need to prove a right angle.</p>
<p>Is it possible to do the problem w/o centroids? I don’t remember centroids mentioned at all in the theory I read beforehand.</p>
<p>I can’t really think of an easier method, the fact that you are given the centroid and not the centre rules out any possibility of finding any useful angles and using standard geometric methods. I doubt this question would ever appear on the SAT.</p>
<p>Well does my inability to prove the two right angles invalidate my entire answer? How did I end up with 1/12 anyway, sheer luck?</p>
<p>Yes and no. If you were give the angles you used as right angles, the point S would be the centre and not the centroid. This would probably make the question much more simplistic and I haven’t really bothered to see how it would pan out, though 1/12 seems a reasonable answer.</p>
<p>haha, Bak0rz’s got it. Yea, this whole centroid question probably and hopefully won’t come up on the real SATs. (although it’s from SAT math practice test).</p>
<p>Here’s another one. This one will is very likely come up on the test.</p>
<p>A telephone company haws run out of seven-digit telephone numbers for a certain area code. To fix the problem, the telephone company will introduce a new area code. What is the maximum number of new seven digit telephone numbers that will be generated for the new area code if both of the following conditions must be met?</p>
<p>condition1 :the first digit cannot be a zero or a one
condition 2: the first three digits cannot be the emergency number 911 or the information number 411.</p>
<p>its a simple problem if you notice that triangle PMN and MNQ and MNR have the same area, QSR’s area is 4 times that of MNS, and PQR is 4 times that of PNM ( similiar triangles). then you let the area of MNS be x, that of SNQ be y and that of PMN be x+y and use the ratios you can solve the problem.</p>
<p>the answer holds for all triangles so you can assume the triangle to be any kind you want.</p>
<p>^Not sure how you got Area of MNR = Area MNQ. It cannot be proven from the given information that MQ = NR. Nor can you assume PQR and PNM are similar. </p>
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A proof does not start with an assumption.</p>
<p>What you effectively said is: Let us assume X = 0, then Y = constant. Therefore, X can be anything.</p>
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<p>Set of numbers not starting with 0 or 1: 8<em>10^6 or 8</em>10<em>10</em>10<em>10</em>10<em>10
Set of numbers with first 3 digits 911 or 411: 2</em>10^4 or 2<em>1</em>1<em>10</em>10<em>10</em>10</p>
<p>Subtraction of sets: 8<em>10^6 - 2</em>10^4 = 7980000</p>
<p>^^i forgot to explain that the 2 triangles have the same base and the same height.</p>
<p>if you let MO perpendicular to PQ and let O be the point on PQ, you will see they share the same height.</p>
<p>they are similiar because N and M are the midpoints of PQ and PR respectively and angle QPR is shared</p>