Question 1209518: (36) PQRS is a rhombus, W is the midpoint of PQ, T is the midpoint of PS and V is the intersection point of QT and SW. What fraction of rhombus PQRS is quadrilateral PWVT?
Link to diagram: https://ibb.co/PsrxgPwG
Found 5 solutions by CPhill, mccravyedwin, greenestamps, Edwin McCravy, ikleyn:
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to solve this problem:
**1. Area of Rhombus:**
Let the side length of the rhombus be 's'. The area of a rhombus can be calculated using its diagonals, but since we're dealing with midpoints and ratios, a simpler approach is to consider it as four congruent triangles. We'll denote the area of rhombus PQRS as [PQRS].
**2. Area of Triangles:**
* **Triangle PQT:** Since W and T are midpoints, PW = WQ = s/2 and PT = TS = s/2. Triangle PQT has half the base and half the height of triangle PQS (which is half of the rhombus). Therefore, [PQT] = (1/2) * (s/2) * (s/2) * sin(∠P) = (1/8) * s² * sin(∠P) = (1/4) * [PQS] = (1/8) * [PQRS].
* **Triangles PSW and QRS:** By the same logic, [PSW] = (1/8) * [PQRS].
**3. Area of Triangle PVW:**
We know that the medians of a triangle divide it into six equal areas. In triangle PQS, QT and SW are medians. Therefore, the area of triangle PVW is 1/6 the area of triangle PQS.
[PVW] = (1/6) * [PQS] = (1/6) * (1/2) [PQRS] = (1/12) * [PQRS]
**4. Area of Quadrilateral PWVT:**
The area of quadrilateral PWVT can be found by subtracting the areas of triangles PVW, PWT, and QWV from the area of triangle PQT.
*Note that PWT and QWV are also 1/6 of the area of the relevant larger triangle, which is 1/4 of the area of the rhombus.*
[PWT] = [QWV] = (1/6) * [PQT] = (1/6) * (1/8) [PQRS] = (1/48) [PQRS]
[PWVT] = [PQT] - [PVW] - [PWT]
[PWVT] = (1/8)[PQRS] - (1/12)[PQRS] - (1/48)[PQRS]
[PWVT] = ((6-4-1)/48) * [PQRS]
[PWVT] = (1/48) * [PQRS]
**5. Fraction of Rhombus:**
The area of quadrilateral PWVT is (1/48) of the area of rhombus PQRS.
Therefore, the fraction of rhombus PQRS that is quadrilateral PWVT is **1/48**.
Answer by mccravyedwin(406)  (Show Source):
You can put this solution on YOUR website!
I went rummaging through the garbage of the last deluge of bad AI "solutions"
and found this interesting rhombus problem buried there. (incorrectly done by
AI.)
Rhombuses are easier to think about when you draw them diamond-shaped, i.e.,
symmetrical with the horizontal and the vertical. So I will draw the figure
that way instead of the way it's drawn on the site of the given link.
As you can see from the second figure below, any rhombus can be partitioned into
8 CONGRUENT isosceles triangles (they might look equilateral but that's not
necessarily the case. They are only isosceles, I just accidentally drew them to
look equilateral.)
Anyway, the second figure below shows that ΔTPW is 1/8 of the rhombus
(area-wise). That's too obvious to bother wasting time to prove.
Quadrilateral PWVT is made up of ΔTPW and ΔTVW. So all that's left is to find
what fraction ΔTVW is of the whole rhombus and add that to 1/8.
 
We will use these two well-known theorems:
Theorem 1: If two triangles are similar, the ratio of their areas is equal to
the square of the ratio of any pair of corresponding sides between them.
Theorem 2: The intersection of the three medians of a triangle, called the
centroid, is located two-thirds of the distance from each vertex to the midpoint
of the opposite side. That is to say, the shorter part of each median from the
centroid is 1/2 the longer part from the centroid. It also says that the
entire median is 3 times the shorter part when divided by the centroid.
Notice that SW, QT and PO are the three medians of ΔSPQ, and V is the centroid
of ΔSPQ.
ΔSVQ is 1/3 of ΔSPQ. Why? Remembering the formula for the area of a triangle,
A=1/2(bh), they have the same base SQ, and by theorem 2, the height OV of ΔSVQ
is 1/3 of the height OP of ΔSPQ.
Since ΔSVQ is 1/3 of ΔSPQ which is 1/2 of the whole rhombus, then ΔSVQ is
(1/3)(1/2) = 1/6 of the whole rhombus.
By theorem 2, VW is 1/2 of SV, and since ΔTVW is similar to ΔSVQ, the area
of ΔTVW, by theorem 1, is (1/2)2=1/4 the area of ΔSVQ
So ΔTVW is (1/4)(1/6) = 1/24th of the whole rhombus.
Therefore quadrilateral PWVT is 1/8 + 1/24 = 3/24 + 1/24 = 4/24 = 1/6 of rhombus PQRS.
Answer: 1/6
Edwin
Answer by greenestamps(13198)  (Show Source):
You can put this solution on YOUR website!
From the statement of the problem, the result is independent of the shape of the rhombus. So choose the "nice" case where it is composed of two equilateral triangles.
In each triangle, the three medians divide the triangle into 6 congruent 30-60-90 right triangles, so quadrilateral PWVT has area 2/6 = 1/3 of the area of that equilateral triangle.
And since the two equilateral triangles are the same size, the area of quadrilateral PWVT is 1/6 of the area of the rhombus.
ANSWER: 1/6
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Comment to Edwin's comment....
If a geometry problem is stated generally so that the desired result is true in all cases, then it is a very powerful problem-solving strategy to solve the problem by selecting a specific case for which the desired result is easily obtained....
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... added later to pacify tutor @ikleyn, who doubts that problems like this can be solved by choosing a "nice" case....
Each diagonal of a rhombus divides the rhombus into two congruent triangles.
In each of those triangles, the three medians divide the triangle into six triangles whose areas are equal; that means the area of quadrilateral PWVT (composed of 2 of the 6 smaller triangles in its triangle) is 2/6 = 1/3 the area of its triangle and thus 1/6 the area of the whole rhombus.
In my original response above, I simply chose to use a rhombus composed of two equilateral triangles to make it easier to see the solution.
Answer by Edwin McCravy(20054)  (Show Source):
You can put this solution on YOUR website!
Comment to greenestamps' solution.
While he is right that the problem does imply that the fraction is
constant for all rhombuses, it is not obvious that it would be.
I do think the student was expected to prove it is constant for all
rhombuses, and not just find it for one particular case.
You might suspect that I did draw the figure as the special case
greenestamps suggested. But I was careful to state that I could
not use that fact to prove it for all rhombuses.
J
Edwin
Answer by ikleyn(52776)  (Show Source):
You can put this solution on YOUR website! .
After reading the @greenestamps comment to Edwin's comment,
I can not leave this subject without my comment.
The point by @greenestamps turns upside down all Euclidian geometry
and all the known Math logic.
Consider one example, this statement, which is one of the many theorems of Euclidian geometry:
in isosceles triangles, the angles at the base are congruent.
This statement relates to all isosceles triangle.
Then, following to the @greenestamps logic, it is enough to check or to prove it
for equilateral triangles, ONLY.
But it is clear to everyone, that this logic does not work.
In all textbooks on Geometry, you will find the proof of the statement valid
for ALL isosceles triangles. No one Geometry textbook says that for proving
general statement, it is enough to check or to prove it for some special/specific case.
Are all these textbooks (including Euclid himself) incorrect and teach us in wrong way ?
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