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Re: ABCD is a quadrilateral, as shown in the diagram below. What is the [#permalink]
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Octobre wrote:
Hi Broall

Thanks for your answer. Your mathematical notation was somehow not well interpreted by the system. Could you please edit if you a have a minute?

Many Thanks


Done. Thank you

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Re: ABCD is a quadrilateral, as shown in the diagram below. What is the [#permalink]
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Abhishiek wrote:
Hey,

What does S mean? (Area)
and how is
\(\frac{S_{AOB}}{S_{BOC}}=\frac{AO}{OC}\)


Yep, S mean the area.

Attachment:
2017-07-24_1034.png
2017-07-24_1034.png [ 9.59 KiB | Viewed 2706 times ]


BH is perpendicular to AC.

We have
\(S_{AOB} = \frac{1}{2} * AO * BH\)
\(S_{BOC} = \frac{1}{2} * OC * BH\)

Hence
\(\frac{S_{AOB}}{S_{BOC}}=\frac{\frac{1}{2} * AO * BH}{\frac{1}{2} * OC * BH}=\frac{AO}{OC}\)

Hope this helps.
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Re: ABCD is a quadrilateral, as shown in the diagram below. What is the [#permalink]
broall wrote:
Bunuel wrote:
ABCD is a quadrilateral, as shown in the diagram below. What is the minimum possible sum of areas of triangles AOD and BOC?


(1) Area of triangles AOB and COD are 16 and 36, respectively.
(2) Area of triangle AOD is equal to the area of triangle BOC.


Attachment:
2017-07-24_1034.png


(1) We have \(S_{AOB} = 16\) and \(S_{COD} = 36\)

Also \(\frac{S_{AOB}}{S_{BOC}}=\frac{AO}{OC}\)
\(\frac{S_{AOD}}{S_{DOC}}=\frac{AO}{OC}\)

Hence \(\frac{S_{AOB}}{S_{BOC}} = \frac{S_{AOD}}{S_{DOC}} \\
\implies S_{AOD} * S_{BOC} = S_{AOB} *S_{DOC} = 16 * 36 \)

ALso \(16 * 36 = S_{AOD} * S_{BOC} \leq \frac{1}{4}( S_{AOD} + S_{BOC})^2 \implies S_{AOD} + S_{BOC} \geq 48\)

Sufficient.

(2) We have no more information about the area of AOD and BOC, hence we can't know the minimum possible sum of areas of triangles AOD and BOC. Insufficient.

The answer is A.



Hey,

What does S mean? (Area)
and how is
\(\frac{S_{AOB}}{S_{BOC}}=\frac{AO}{OC}\)
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Re: ABCD is a quadrilateral, as shown in the diagram below. What is the [#permalink]
broall wrote:
Bunuel wrote:
ABCD is a quadrilateral, as shown in the diagram below. What is the minimum possible sum of areas of triangles AOD and BOC?

(1) Area of triangles AOB and COD are 16 and 36, respectively.
(2) Area of triangle AOD is equal to the area of triangle BOC.



ALso \(16 * 36 = S_{AOD} * S_{BOC} \leq \frac{1}{4}( S_{AOD} + S_{BOC})^2 \implies S_{AOD} + S_{BOC} \geq 48\)



Can you explain how you came up with the above part?
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Re: ABCD is a quadrilateral, as shown in the diagram below. What is the [#permalink]
ALso 16∗36=SAOD∗SBOC≤14(SAOD+SBOC)2⟹SAOD+SBOC≥48

Can you please explain this
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Re: ABCD is a quadrilateral, as shown in the diagram below. What is the [#permalink]
broall wrote:
Bunuel wrote:
ABCD is a quadrilateral, as shown in the diagram below. What is the minimum possible sum of areas of triangles AOD and BOC?


(1) Area of triangles AOB and COD are 16 and 36, respectively.
(2) Area of triangle AOD is equal to the area of triangle BOC.


Attachment:
2017-07-24_1034.png


(1) We have \(S_{AOB} = 16\) and \(S_{COD} = 36\)

Also \(\frac{S_{AOB}}{S_{BOC}}=\frac{AO}{OC}\)
\(\frac{S_{AOD}}{S_{DOC}}=\frac{AO}{OC}\)

Hence \(\frac{S_{AOB}}{S_{BOC}} = \frac{S_{AOD}}{S_{DOC}} \\
\implies S_{AOD} * S_{BOC} = S_{AOB} *S_{DOC} = 16 * 36 \)

ALso \(16 * 36 = S_{AOD} * S_{BOC} \leq \frac{1}{4}( S_{AOD} + S_{BOC})^2 \implies S_{AOD} + S_{BOC} \geq 48\)

Sufficient.

(2) We have no more information about the area of AOD and BOC, hence we can't know the minimum possible sum of areas of triangles AOD and BOC. Insufficient.

The answer is A.



Although the answer is correct, but I feel there is one mistake in the formula used:

\((a+b)^2\)=\(a^2\)+\(b^2\)+\(2ab\)
=>\((a+b)^2\) > \(2ab\) {As squared term is always greater than 0}
=>\(ab\)<\(\frac{1}{2} (a+b)^2\)

So instead of \(\frac{1}{4}\), it should be \(\frac{1}{2}\) from which we get \((a+b)>24\sqrt{2}\).

Thus the minimum value as required is \(24\sqrt{2}\).
Please correct me if I am wrong. Thanks.
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Re: ABCD is a quadrilateral, as shown in the diagram below. What is the [#permalink]
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Re: ABCD is a quadrilateral, as shown in the diagram below. What is the [#permalink]
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