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Intersection And Maximum Triangle (Posted on 2008-03-30) Difficulty: 3 of 5
A tangent to the ellipse x2/9 + y2/2 = 1 intersects the circle x2 + y2 = 9 at the points P and Q. It is known that R is a point on the circle x2+ y2 = 9. Each of P, Q and R are located above the x-axis.

For example, the coordinates of R cannot be (3,0), since (3,0) is not located above the x axis.

Determine the maximum area of the triangle PQR.

See The Solution Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution re: Well..... correction | Comment 4 of 5 |
(In reply to Well..... by Charlie)

The corrected original table is:

 y-coord of
pt of tangency

R    \    0.100000  0.319036  0.538071  0.757107  0.976142  1.195178  1.414213

0.0000   2.3372933 6.3651187 8.1840718 8.3008114 7.5093707 6.2393841 3.7437402
0.0750   2.3325406 6.3255642 8.0970489 8.1718618 7.3489020 6.0564763 3.5453082
0.1500   2.3270576 6.2840444 8.0075657 8.0405342 7.1864607 5.8722086 3.3468749
0.2250   2.3208423 6.2405538 7.9156152 7.9068217 7.0220414 5.6865771 3.1484403
0.3000   2.3138911 6.1950831 7.8211859 7.7707132 6.8556347 5.4995754 2.9500044
0.3750   2.3061993 6.1476191 7.7242613 7.6321926 6.6872274 5.3111943 2.7515672
0.4500   2.2977603 6.0981447 7.6248199 7.4912393 6.5168024 5.1214222 2.5531286
0.5250   2.2885665 6.0466386 7.5228350 7.3478276 6.3443382 4.9302441 2.3546887
0.6000   2.2786081 5.9930753 7.4182748 7.2019265 6.1698091 4.7376424 2.1562474
0.6750   2.2678741 5.9374245 7.3111012 7.0534994 5.9931850 4.5435962 1.9578048
0.7500   2.2563513 5.8796510 7.2012704 6.9025037 5.8144303 4.3480812 1.7593607
0.8250   2.2440246 5.8197143 7.0887315 6.7488904 5.6335046 4.1510693 1.5609152
0.9000   2.2308770 5.7575682 6.9734267 6.5926036 5.4503613 3.9525285 1.3624682
0.9750   2.2168888 5.6931601 6.8552901 6.4335795 5.2649477 3.7524224 1.1640198
1.0500   2.2020379 5.6264304 6.7342472 6.2717462 5.0772039 3.5507098 0.9655698
1.1250   2.1862993 5.5573117 6.6102136 6.1070221 4.8870624 3.3473441 0.7671182
1.2000   2.1696446 5.4857280 6.4830941 5.9393152 4.6944469 3.1422726 0.5686650
1.2750   2.1520421 5.4115935 6.3527812 5.7685215 4.4992711 2.9354360 0.3702101
1.3500   2.1334557 5.3348110 6.2191535 5.5945238 4.3014378 2.7267671 0.1717534
1.4250   2.1138445 5.2552709 6.0820735 5.4171890 4.1008366 2.5161899 0.0267051
1.5000   2.0931622 5.1728485 5.9413853 5.2363666 3.8973425 2.3036182 0.2251655
1.5750   2.0713563 5.0874019 5.7969112 5.0518847 3.6908133 2.0889538 0.4236279
1.6500   2.0483663 4.9987692 5.6484484 4.8635473 3.4810861 1.8720847 0.6220924
1.7250   2.0241232 4.9067640 5.4957638 4.6711291 3.2679741 1.6528819 0.8205592
1.8000   1.9985469 4.8111713 5.3385880 4.4743697 3.0512614 1.4311966 1.0190283
1.8750   1.9715445 4.7117408 5.1766077 4.2729663 2.8306968 1.2068555 1.2175000
1.9500   1.9430070 4.6081790 5.0094556 4.0665638 2.6059857 0.9796557 1.4159745
2.0250   1.9128051 4.5001383 4.8366964 3.8547415 2.3767793 0.7493565 1.6144518
2.1000   1.8807843 4.3872026 4.6578093 3.6369960 2.1426600 0.5156702 1.8129325
2.1750   1.8467570 4.2688673 4.4721626 3.4127165 1.9031212 0.2782473 2.0114167
2.2500   1.8104921 4.1445104 4.2789776 3.1811507 1.6575388 0.0366576 2.2099049
2.3250   1.7716996 4.0133516 4.0772774 2.9413540 1.4051293 0.2096389 2.4083976
2.4000   1.7300071 3.8743884 3.8658072 2.6921136 1.1448870 0.4613359 2.6068955
2.4750   1.6849228 3.7262974 3.6429101 2.4318279 0.8754833 0.7193492 2.8053995
2.5500   1.6357734 3.5672670 3.4063182 2.1583048 0.5951000 0.9849324 3.0039107
2.6250   1.5815937 3.3946998 3.1527798 1.8684012 0.3011302 1.2598829 3.2024308
2.7000   1.5209091 3.2046269 2.8773266 1.5573147 0.0104096 1.5469470 3.4009626
2.7750   1.4512491 2.9904005 2.5716361 1.2170009 0.3461915 1.8507250 3.5995104
2.8500   1.3678228 2.7391277 2.2195682 0.8318586 0.7191558 2.1801384 3.7980828
2.9250   1.2583359 2.4177232 1.7797040 0.3618523 1.1625094 2.5580819 3.9967016

When homed in on the apparent max:

          0.662000  0.662050  0.662100  0.662150  0.662200  0.662250  0.662300
0.0000   8.4024596 8.4024597 8.4024598 8.4024598 8.4024598 8.4024597 8.4024595
0.0003   8.4020905 8.4020906 8.4020906 8.4020906 8.4020905 8.4020904 8.4020902
0.0005   8.4017213 8.4017214 8.4017214 8.4017213 8.4017212 8.4017211 8.4017208
0.0008   8.4013521 8.4013522 8.4013521 8.4013521 8.4013519 8.4013517 8.4013514
0.0010   8.4009829 8.4009829 8.4009829 8.4009827 8.4009826 8.4009823 8.4009820
0.0013   8.4006136 8.4006136 8.4006135 8.4006134 8.4006132 8.4006129 8.4006126
0.0015   8.4002444 8.4002443 8.4002442 8.4002440 8.4002438 8.4002435 8.4002431

So it seems the maximum area approaches about 8.4024598 when the point of tangency on the ellipse has y-coordinate of about 0.66215 and point R approaches the x-axis on the opposite quadrant (1 vs 2) from the point of tangency.

Peering into the variables of the program, the point of tangency would be at (-2.65085,0.66215) and R would be approaching as close to (3,0) as you dare to get. The intersection points of the tangent line and the circle would be at (-2.9768,0.37215) and (-0.023102,2.999911).

So strictly speaking there'd be no maximum, since the interval is open ended (does not include it's endpoint at (3,0)).

DEFDBL A-Z
CLS

minE = .1#: maxE = SQR(2#)
minC = 0#: maxC = 3#

FOR ey0 = minE TO maxE STEP (maxE - minE) / 6.000001#
 ct = ct + 1
 LOCATE 1, INT(ct * 10)
 PRINT USING "##.######"; ey0;
NEXT

ctc = 0
FOR cy0 = minC TO maxC STEP (maxC - minC) / 40
 ctc = ctc + 1
 cx0 = SQR(9 - cy0 * cy0)
 LOCATE ctc + 1, 1
 PRINT USING "#.####"; cy0
 ct = 0
 FOR ey0 = minE TO maxE STEP (maxE - minE) / 6.000001#
   ex0 = -SQR(9 * (1 - ey0 * ey0 / 2))
   slope = (-4 * ex0 / 9) / (2 * SQR(2 - 2 * ex0 * ex0 / 9))
   b = ey0 - slope * ex0
   m = slope
   x1 = (-2 * b * m - SQR(4 * b * b * m * m - 4 * (m * m + 1) * (b * b - 9))) / (2 * (m * m + 1))
   x2 = (-2 * b * m + SQR(4 * b * b * m * m - 4 * (m * m + 1) * (b * b - 9))) / (2 * (m * m + 1))
   y1 = m * x1 + b
   y2 = m * x2 + b
   a = SQR((x1 - x2) ^ 2 + (y1 - y2) ^ 2)
   b = SQR((x1 - cx0) ^ 2 + (y1 - cy0) ^ 2)
   c = SQR((x2 - cx0) ^ 2 + (y2 - cy0) ^ 2)
   s = (a + b + c) / 2
   area = SQR(s * (s - a) * (s - b) * (s - c))
   ct = ct + 1
   LOCATE ctc + 1, INT(ct * 10)
   PRINT USING "#.#######"; area
 NEXT
NEXT

The changed numerator is bolded and underlined in the revised program listing above.

 

Edited on March 30, 2008, 5:39 pm
  Posted by Charlie on 2008-03-30 17:32:36

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