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In the link below, I found explanation showing that a triangle formed by connecting midpoints of the triangle divides the area of this bigger triangle into 4 equal areas: http://mathworld.wolfram.com/MedialTriangle.html

M, N, and O are midpoints of sides AB, BC, and AC of triangle ABC. What is the area of triangle MON?

The main idea here is realizing that \(S_{MON}=\frac14*S_{ABC}\)

1. Let's consider vertex A: M and O are midpoints of AB and AC. In other words, all linear sizes of MAO triangle is smaller by 2 times than all linear sizes of BAC. Therefore,\(S_{MAO}=\frac14*S_{ABC}\)

2. Applying the same reasoning for each vertex we will get: \(S_{MON}=S_{ABC} - (S_{MAO}+S_{MBN}+S_{NCO}) = S_{ABC} - (\frac14*S_{ABC}+\frac14*S_{ABC}+\frac14*S_{ABC}) =\frac14*S_{ABC}\)
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M, N, and O are midpoints of sides AB, BC, and AC of triangle ABC. What is the area of triangle MON?

The main idea here is realizing that \(S_{MON}=\frac14*S_{ABC}\)

But how do you arrive at this conclusion, were are not told that either ABC or MNO are equilateral. AB, BC, and OC can each be a different length, as can MN, ON, OM.

M, N, and O are midpoints of sides AB, BC, and AC of triangle ABC. What is the area of triangle MON?

The main idea here is realizing that \(S_{MON}=\frac14*S_{ABC}\)

But how do you arrive at this conclusion, were are not told that either ABC or MNO are equilateral. AB, BC, and OC can each be a different length, as can MN, ON, OM.

Mid point therom says that a triangle made from connecting mid-points of the sides of a given triangle is 1/4 of the original triangle. In that case, it is not required to be equilateral.
_________________

for any triangle, ABC if you join mid points and divide it into 4 parts, the area will be divided into 4 triangles with each having an area equal to 1/4 of triangle ABC.

Lets prove is with simple method, visualize the triangle ABC on co-ordinate plane

A = (0,0) B = (x,0) C = (a,b)

Area = xb/2 (Note that area does not depend on a)

M = (x/2, 0) N = ((x+a)/2, b/2) O = (a/2, b/2)

base of triangle, NO = x/2 Height = b/2 Area of MNO = xb/8

Similarly triangle AMO base AM = x/2 height = b/2 Area = xb/8

But how do you arrive at this conclusion, were are not told that either ABC or MNO are equilateral. AB, BC, and OC can each be a different length, as can MN, ON, OM.

This problem test similarity. ABC and MAO (as other small triangles) are similar triangles: the same angle A and the same relation between AB/AC=AM/AO.
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MN, NO and OM each are midsegments of triangle ABC (midsegment is a line segment joining the midpoints of two sides of a triangle). Important property of a midsegment: the midsegment is always half the length of the third side. So, \(MN=\frac{AC}{2}\), \(NO=\frac{AB}{2}\) and \(OM=\frac{BC}{2}\)

Next, since each side of triangle MNO is half of the side of triangle ABC then these triangles are similar (the ratio of all the sides are the same). Important property of similar triangles: if two similar triangles have sides in the ratio \(\frac{x}{y}\), then their areas are in the ratio \(\frac{x^2}{y^2}\).

Since the sides of two similar triangles MNO and ABC are in the ratio 1:2 then then their areas are in the ratio 1:4 --> (area of MNO)=(area of ABC)/4.

So, in order to find the area of MNO we should find the area of ABC.

(1) The area of ABC is \(\frac{\sqrt{3}}{4}\). Sufficient. (2) ABC is an equilateral triangle with height \(\frac{sqrt3}{2}\) --> we can find the area of equilateral triangle with given altitude. Sufficient.

Based on the prompt, We know that the each side of triangle MON is going to be exactly HALF the length of the corresponding sides of triangle ABC. Which means that these two are SIMILAR triangles, and we will be able to figure out the area of MON if we are given the area of ABC. based on the ratio s^2:s^2 (s=side).

1) SUFFICIENT. 2) if ABC is an equilateral triangle, its height is the same from any base, and also cuts the triangle ABC in half to make it 2 right triangles with sides ratio of x:x^(1/2):2x. Knowing the side of the longest leg, we are able to calculate the rest of the sides and hence the area of ABC. SUFFICIENT.