Kite Riddle II

Geometry Level 1

A kite is divided into four regions by two line segments that connect the midpoints of opposite sides. Is the total blue area equal to the total orange area?

Yes, always No, not necessarily No, never

5 solutions

Chew-Seong Cheong
Nov 19, 2017

Draw lines from the vertices of the kite to the center as shown in the figure above. Then the blue area $A_{\color{#3D99F6}\text{blue}} = 2(a+b)$ and $A_{\color{#EC7300}\text{orange}} = 2(c+d)$ . Now note that triangle $a$ and triangle $d$ have the same base length and same height $h_1$ , therefore $a=d$ . Similarly, $b=c$ and therefore $A_{\color{#3D99F6}\text{blue}} = 2(a+b) = 2(c+d) = A_{\color{#EC7300}\text{orange}}$ .

Yes, always $\boxed{A_{\color{#3D99F6}\text{blue}} = A_{\color{#EC7300}\text{orange}}}$ .

Nice and easy.Thanks.

Ahsan Kabir Pranto - 3 years, 6 months ago

Hi, it might be a dumb question but I am not sure how you know that the line segments seperating d and a (in your example) is the height. (Height implies it is perpendicular and creates a 90degree angle. But might it also be slightly slanted right?-(thus not being h and making the triangles different).

Same question applies to the line segment seperating b and c.

Alex A - 3 years, 6 months ago

The line segments dividing $a$ and $d$ , and $b$ and $c$ are not the heights. The heights are shown in the figure as $h_1$ for triangles $a$ and $d$ (which I have also mentioned in the solution) and $h_2$ for triangles $b$ and $c$ . Both $h_1$ and $h_2$ are perpendicular to the respective sides also shown in the figure.

Chew-Seong Cheong - 3 years, 6 months ago

Sorry but I am still unconvinced. “Now note that triangle a and triangle d have the same base and same height h1, therefore a = d.”

This is what you said in your solution. Triangles a and d have the same line segment seperating them but that line is not h1 by the looks of the image. h1 (90 degrees) is a line that divides “a” into 2 parts.

You admit in your response to my first comment that the line seperating a and d is not the height. So then how do you know the line seperating a and d divides them into equal areas? (Shouldn’t this only be true when the line dividing the areas is perpendicular to the base..?)

Alex A - 3 years, 6 months ago

@Alex A – Sorry, I mean the same base length. Both $a$ and $d$ has the same base length of $b$ the two line segments that marked by $|$ as equal length. And they have the same height of $h_1$ , then their area are both $\dfrac {bh_1}2$ .

Sorry, I couldn't attach a figure to show you though I tried. But the solution is correct because there are more than 8700 solvers despite you are unconvinced. Perhaps it is easier to see if you don't the doubt the solution of thousands of solvers. Just discuss with a friend.

Chew-Seong Cheong - 3 years, 6 months ago

@Chew-Seong Cheong – Oh I see, thanks for the clarification. I was just confused by the line that separated both areas so that’s why I was still unconvinced.

Alex A - 3 years, 6 months ago

It doesn't actually matter what the heights of the triangles are. What matters is that a & d have the same height (the distance from the base line to the central point). Likewise for b & c.

Jerry Barrington - 3 years, 6 months ago

Hello, I wanted to check, I bisected the kite into two equilateral triangles where | meets || and concluded that line segments straight through the middle subdividing the triangles each are an equal ratio, therefore blue and orange will be equal for all cases. Is this sound logic, or will it fall short in some cases?

Patrick Bamber - 3 years, 6 months ago

Sorry, I don't quite understand what you mean.

Chew-Seong Cheong - 3 years, 6 months ago

Exactly. It took me a while to realise that the kite can be dissected like this. As Euclid taught us, triangles on equal bases in the same parallels are equal.

Stewart Gordon - 3 years, 6 months ago

It seems to me, that answer is not correct otherwise I misunderstood the task. I assumed, that segments marked by a single trait are equal and segments marked by double trait are equal also. Let's name a single trait segment as "a" and double trait segment as "b". Than the area A(blue)=a b+a b=2 a b. The orange area A(orange)=a a+b b=a^2+b^2. Than A(blue) < A(orange). It works for any values for "a" and "b".

Denis Tolstov - 3 years, 6 months ago

The answer cannot be wrong because there are 9,667 solvers. Discuss the problem with your friends.

Chew-Seong Cheong - 3 years, 6 months ago

Number of solvers doesn't determine that the answer is right. The geometric logic does.

Jerry Barrington - 3 years, 6 months ago

@Jerry Barrington – Yes, of course. But it is a good indication that the solution posted is correct. I just wanted those who inquired to look at the solution carefully before asking. I feel some fail to see the solution because they doubt solutions by others.

Chew-Seong Cheong - 3 years, 6 months ago

That formula for area works only if the regions are rectangular, which they aren't in this instance (except in the degenerate case where $a = b$ , but then $2ab = a^2 + b^2$ ).

Stewart Gordon - 3 years, 6 months ago

We note that $\triangle a$ and $\triangle d$ has the same base length; let it be $b_a$ , then the area of $\triangle a$ is $\frac 12 b_ah_1$ , the area of $\triangle d$ is also $\frac 12 b_ah_1$ .

Chew-Seong Cheong - 3 years, 6 months ago

Venkatachalam J
Nov 19, 2017

Let draw red rectangle connecting the midpoints of the kite and extended parallel lines of the rectangle as in the above diagram.

Clearly, in the red rectangle the area of blue and orange are equal(each of the small rectangle divided into half blue and half orange). Also, outside to the red rectangle the area of blue and orange are equal.(extended kite sides intersecting the parallel lines-angles of intersection & side length are equal). The equal blue and orange area are indicated with arrow marks.

Hence the area of blue and orange is always equal.

Moderator note:

This trick is nearly identical to the one used last week in a parallelogram puzzle .

Sorry, i’m probably slow but I don’t see how the area outside the red rectangle is equal.?

Tyvon O - Addo - 3 years, 6 months ago

The red rectangle is extended and there are three red parallel lines and the black line is intersecting the red line. Since it is intersecting the parallel lines the angles are equal (triangles has 90 degree and two equal angles). Since the marking arrows clearly show which blue and orange parts of the triangles are equal.

Venkatachalam J - 3 years, 6 months ago

Worranat Pakornrat
Nov 4, 2017

By drawing the horizontal lines joining the kite's midpoints and the major diagonal, all of them are parallel due to similarity. Then it's obvious to see that the red regions are in fact smaller kites with half of the original minor diagonal due to the midpoint ratio. Thus, the sum of the red kites result in half of the one big kite, making the green and red areas equal.

The solution could be clarified. At first glance, I thought you were claiming that the red kites are similar.

Essentially, what you're doing here is
1. Consider the top half of the kite
2. The red triangles have area equal to 1/2 * base length * 1/2 height.
3. Hence, the red triangles form half of the top half.

Calvin Lin Staff - 3 years, 7 months ago

Doesn't cutting through the midpoints mean cutting the figure in half from the midpoint of one edge to the midpoint of the opposite edge? While the amount of green would be the same on each half, the amount of red on one half would be greater than the amount of red on the other half.

Gregory Lewis - 3 years, 7 months ago

I'm not saying the red kites are similar though; the red ones do have different areas. However, the sum of 2 kites = sum of 2 green areas = half the big kite.

Worranat Pakornrat - 3 years, 7 months ago

Ok, I see where the confusion is. You don't mean to cut an already painted figure, you mean to take an unpainted figure and color it by connecting the midpoints. I was having a hard time guessing what the question was asking.

Gregory Lewis - 3 years, 7 months ago

I don't think this question is appropriate in the 'basic' section of problems of the week

Aaron Zolnai-Lucas - 3 years, 6 months ago

Best explanation

Terry Smith - 3 years, 6 months ago

Since the kite is symmetrical about x- axis, one should noticed that for the 1/2 of the kite ( 1 blue area = 2 halfs of orange areas) means one blue area = one orange area in the upper half and so for the complete kite..

Adnan Alzahrani - 3 years, 6 months ago

Bauet Stevenson
Nov 25, 2017

I simply think you can draw the diagonals and as they intersect at 90degrees they also intersect theline segments dividing all equally to prove the equality of the sides.

Mike Waddingham
Nov 24, 2017

1.If you pull the 2 extreme Orange corners together, all 4 segments close up together to nothing. If you pull the 2 blue corners together the same thing happens. This can only be so if the shape is symmetrical.

Alternatively, if the kite were a cake and you wanted to cut 4 equal portions, you would instinctively cut from mid point to mid point. Moving a cut away from the mid point, towards an edge, it becomes increasingly obvious that the result will be wrong. And if you draw the kite with a very blunt head, it becomes obvious that corner to corner does't work either. Blueskies.

Kite Riddle II

5 solutions

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