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Question Bank - Geometry - Hemant Malhotra

Smitirupa Mahapatra last edited by

@hemant_malhotra Ans 7.4?

r= sqrt(6.75^2+3.05^2)= 7.41 approx.
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Smitirupa Mahapatra last edited by @hemant_malhotra

@hemant_malhotra Ans 26 ?
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Smitirupa Mahapatra last edited by @hemant_malhotra

@hemant_malhotra Sir, 49? Two triangles can be formed, one with ABC 24 area and another with ACD 25 area.
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Tulip Roy last edited by @hemant_malhotra

@hemant_malhotra By similarity d.5/4
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Tulip Roy last edited by @hemant_malhotra

@hemant_malhotra b. 18rt2
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prat77ju last edited by @hemant_malhotra

@hemant_malhotra a+p = 2l+2b+lb = (l+2)(b+2)-4 = 100/102/104/106/108 only 102+4 = 106= 53*2 not possible . so 102 ?
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sudheer last edited by sudheer @hemant_malhotra

@hemant_malhotra Is 43 the maximum perimeter possible ?
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sudheer last edited by @hemant_malhotra

@hemant_malhotra using mass point geometry i got answe as 1/2 (144)=72 ?
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Harsha Vardhan Reddy last edited by @hemant_malhotra

@hemant_malhotra 50 intersection points ??
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Harsha Vardhan Reddy last edited by @hemant_malhotra

@hemant_malhotra 7.4 is the answer !!
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@raman_sharma BC =1 , CA =2 so AB =rt5using similarity in triangle ABC and ADCwe get CD= 2/rt5 and diameter of circle as 4/rt5 now using tangent secant theoremCD^2 = CE * AC4/5 = CE *2CE =0.4so AE=1.6hence 1.6/0.4 = 4
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@hemant_malhotra said in Question Bank - Algebra - Hemant Malhotra: Let a < b < c be the three real roots of the equation √2014 x^3 - 4029 x^2 + 2 = 0. Find b(a + c). https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_9
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@mbatious CDBA suits better?
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The roots are a and b: a + b = p and ab = 12 (a + b)^2 = p^2 (a - b)^2 = (a + b)^2 - 4ab => (a - b)^2 = p^2 - 12 * 4 = p^2 - 48 If |a - b| ≥ 12 { Difference between the roots is at least 12} then, (a - b)^2 ≥ 144 p^2 - 48 ≥ 144 p^2 ≥ 192 P ≥ 8√3 or P ≤ -8√3
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@sumit-agarwal In such scenarios like finding distinct values of [x^2/n] where x can be from 1, 2, 3 ... n[1^2/n], [2^2/n] ... [(n/2)^2/n] will yield all numbers from 0 to [n/4] (means [n/4] + 1 distinct integers)Then the next set (from [(n/2 + 1)^2/n] till [n^2/n] will be all different integers (means [n/2] distinct integers)So the number of distinct integers would be [n/2] + [n/4] + 1 if n = 100,number of distinct integers would be [100/2] + [100/4] + 1 = 76 if n = 2014,number of distinct integers would be [2014/2] + [2014/4] + 1 = 1511 if n = 13number of distinct integers would be [13/2] + [13/4] + 1 = 10 Just trying to generalize a solution shared by Kamal sir (Quant Boosters - Set 1 - Q2). You can try out with various numbers (may be smaller numbers) so that this can be verified.
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@vikas_saini can u plz explain 2nd and 3rd+4th step again. 2nd step:what is -1-what are we reducing in it 3rd step: 3rd+4th step doesnt giv value in 2nd step. what are we missing in it. 3rd+4th= 2T-2

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