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805+ (Hard)|   Geometry|      
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Bunuel
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In a figure above, triangle PQR is right-angled at angle R. If QR = 4 03 May 2020, 10:51
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C
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42% (02:48) correct 58% (03:05) wrong based on 77 sessions

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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. \(\frac{QT}{TR} >1\)

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C
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In a figure above, triangle PQR is right-angled at angle R. If QR = 4 Updated on: 03 May 2020, 13:24
Originally posted by firas92 on 03 May 2020, 12:18.
Last edited by firas92 on 03 May 2020, 13:24, edited 1 time in total.
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Bunuel

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. \(\frac{QT}{TR} >1\)

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In a figure above, triangle PQR is right-angled at angle R. If QR = 4 03 May 2020, 13:05
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Bunuel

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

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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
Screenshot 2020-05-04 at 8.53.31 PM.png [ 24.95 KiB | Viewed 5124 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
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In a figure above, triangle PQR is right-angled at angle R. If QR = 4 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).
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In a figure above, triangle PQR is right-angled at angle R. If QR = 4 18 Aug 2020, 08:50
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lacktutor


One of features of ∆PRQ is \(b+a > 4\)
---> \( \frac{8}{(x-2)} >4\)
\(\frac{2}{(x-2)} > 1\)
\(\frac{(x-4)}{( x-2)} < 0\)

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lacktutor How did you get from \(\frac{2}{(x-2)} > 1\) to \(\frac{(x-4)}{( x-2)} < 0\)? Which substitution or identity allowed that? Of course it's true intuitively, since x > 0 must be true, but why does it even matter?
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In a figure above, triangle PQR is right-angled at angle R. If QR = 4 19 Aug 2020, 17:03
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Simple solution


There's a much easier and faster method of solving this. This involves picking values for TR that are suggested by our answer choices, instead of trying to work out the answer and pretending we don't already have information in the answer choices.

Let's call TR = x. Then QT = 4-x. We want to set bounds on x so that we can see a relationship in the fraction (4-x)/x.

- Can x be 2? If x = 2, then 4-x=2. For that to be true and for the relationship PR + TR = QT + PQ to also be true, then PR = PQ and the triangle is isosceles. HOWEVER, since we know that angle PRQ is 90 degrees, the triangle isn't isosceles. D is wrong, as TR and QT are not equal.

- Can x be less than 2? If TR=x<2, then for PR + TR = QT + PQ to be true, since TR<QT, PR>PQ. This also cannot be true, since PR is the height of a right triangle and PQ is the hypotenuse. E is wrong, as TR>2>QT.

Therefore x>2. A simpler way of saying this -- since by definition the hypotenuse PQ > height PR, for PR + TR = QT + PQ to be true, then TR > QT. Therefore TR=x>2.

- Next, can QT/TR = 1/3? In this case, QT=1 and TR=3. That's not possible, since TR is the base of a right triangle, and its hypotenuse PT is equal to 3. Therefore TR must be less than 3. B is wrong.

A or C? We just said that TR < 3, so QT>1 therefore QT/TR > 1/3. C is the correct answer.

TLDR;


TR = 4 - QT. By definition, the hypotenuse PQ > height PR, so for PR + TR = QT + PQ to be true, then TR > QT. Therefore TR>2 and QT<2. Then, TR cannot be 3 or larger since PT=3. So the bounds on TR are: 2 < TR < 3. So 1/3 < QT/TR < 1.

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In a figure above, triangle PQR is right-angled at angle R. If QR = 4 06 Mar 2025, 05:37
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