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03 May 2020, 10:51
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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42% (02:48) correct
58%
(03:05)
wrong
based on 77
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In a figure above, triangle PQR is right-angled at angle R. If the perimeter of triangle PQT equals to the perimeter of triangle PTR, then:
A. \(0 < \frac{QT}{TR} < \frac{1}{3}\)
B. \(\frac{QT}{TR} = \frac{1}{3}\)
C. \(\frac{1}{3} < \frac{QT}{TR} < 1\)
D. \(\frac{QT}{TR} = 1\)
E. \(\frac{QT}{TR} >1\)
Are You Up For the Challenge: 700 Level Questions
Attachment:
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firas92
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03 May 2020, 13:05
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Bunuel
Given: In a figure above, triangle PQR is right-angled at angle R.
Asked: If the perimeter of triangle PQT equals to the perimeter of triangle PTR, then:
Attachment:
Screenshot 2020-05-04 at 8.53.31 PM.png [ 24.95 KiB | Viewed 5276 times ]
Perimeter of triangle PTR = PR + RT + PT = PR + RT + 3
Perimeter of triangle PQT = PQ + TQ + PT = PQ + TQ + 3
Since the perimeter of triangle PQT equals to the perimeter of triangle PTR
PR + RT = PQ + TQ (1)
Let TR be x and and PR be y
PQ^2 = 4^2 + y^2 = 16 + y^2
Since x^2 + y^2 = 9
PQ^2 = 16 + 9 - x^2 = 25 - x^2
TQ = 4 - x
TQ + PQ = x + y = TQ + PQ = \((4-x) + \sqrt{25 - x^2}\)
\((2x + y - 4)^2 = 25 - x^2\)
\(4(x-2)^2 + 4(x-2)y + y^2 = 25 - x^2 \)
\(4x^2 - 16x + 16 + 4xy - 8y = 25 - 9 = 16\)
\(x^2 - 4x + xy - 2y = 0 \)
\(y = \frac{(x^2 - 4x)}{(2 -x)} >0\)
\(y = \frac{x(4-x)}{(x-2)} > 0 ; x>2\)
\(y^2 = 9 - x^2 = \frac{x^2 (4-x)^2}{(x-2)^2} = (3-x)(3+x) > 0 ; x < 3\)
2<x<3
\(\frac{QT}{TR} = \frac{(4-x)}{x} = \frac{4}{x} - 1\)
\(For\ x = 2 ; \frac{(4-x)}{x} = \frac{4}{2} - 1 = 1\)
\(For\ x = 3; \frac{(4-x)}{x} = \frac{4}{3} - 1 = \frac{1}{3}\)
\(\frac{1}{3} < \frac{QT}{TR} < 1\)
IMO C
06 May 2020, 22:03
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Please See The Attached File
Since ∆ PQR is a right-angled triangle, \(a^{2}+ 16= b^{2}\) ---> \((b-a)(b+a )= 16\)
The perimeter of triangle PQT equals to the perimeter of triangle PTR:
---> \(a+x +3 = b+ (4-x) + 3\)
\(b -a = 2x -4 \)
---> \((b+a )= \frac{16}{(2x -4)}= \frac{8}{(x-2)}\)
One of features of ∆PRQ is \(b+a > 4\)
---> \( \frac{8}{(x-2)} >4\)
\(\frac{2}{(x-2)} > 1\)
\(\frac{(x-4)}{( x-2)} < 0\)
Since ∆ PTR is a right-angled triangle, \(x <3\).
So, we got x ----> \(x> 2\) and \(x <3 \)
Well,
\(x(2)= \frac{4-x }{x} = \frac{4-2}{2}= 1\)
\(x(3) =\frac{4-x }{x} = \frac{4-3}{3}= \frac{1}{3} \)
\(\frac{1}{3} < \frac{4-x}{x} < 1\)
Answer (C).
Since ∆ PQR is a right-angled triangle, \(a^{2}+ 16= b^{2}\) ---> \((b-a)(b+a )= 16\)
The perimeter of triangle PQT equals to the perimeter of triangle PTR:
---> \(a+x +3 = b+ (4-x) + 3\)
\(b -a = 2x -4 \)
---> \((b+a )= \frac{16}{(2x -4)}= \frac{8}{(x-2)}\)
One of features of ∆PRQ is \(b+a > 4\)
---> \( \frac{8}{(x-2)} >4\)
\(\frac{2}{(x-2)} > 1\)
\(\frac{(x-4)}{( x-2)} < 0\)
Since ∆ PTR is a right-angled triangle, \(x <3\).
So, we got x ----> \(x> 2\) and \(x <3 \)
Well,
\(x(2)= \frac{4-x }{x} = \frac{4-2}{2}= 1\)
\(x(3) =\frac{4-x }{x} = \frac{4-3}{3}= \frac{1}{3} \)
\(\frac{1}{3} < \frac{4-x}{x} < 1\)
Answer (C).
Attachments
right_angled.png [ 10.51 KiB | Viewed 5192 times ]
