# Why would the triangles join up to a rhombus?

The question I am going to present may as well sound very dumb. But this is becoming a hell of a confusing thing for me.

The question is from ISI B.Math-B.Stat entrance exam 2022 UGA question paper. It is problem 29. The question included pictures, but I am going to rephrase the question in a way such that the pictures are not needed but I will still include the pictures.

If $$\triangle{APB}$$ has area $$4$$, $$\triangle{BPC}$$ has area $$5$$, $$\triangle{CPD}$$ has area $$x$$ and $$\triangle{APD}$$ has area $$13$$, where $$AB=BC=CD=DA=6$$

Then find the value of $$x$$.

Here is the picture:

Now. Here's my question. Apparently if we join up the triangles we get a rhombus. Like this:

Then we can simply draw two perpendicular going through the point $$P$$. Then considering the areas of the triangles we will get that

$$\triangle{APB}+\triangle{CPD}=\triangle{BPC}+\triangle{APD}$$

Meaning that, $$x=13+5-4=14$$

But why would the the triangle add up to a rhombus. I mean it intuitively makes sense but can we give a logical reason. I mean it could've been the case that when we join up the triangles then two of the triangles just don't meet. Like this:

Now, after we connect $$C$$ and $$E$$ (shown by the dotted line), if we can show that $$CE=6$$ then we can simply show that such a picture is a contradiction because then all the sides of $$\triangle{PCE}$$ and $$\triangle{PCD}$$ will be equal which will mean that they are similar triangles but clearly $$\angle{PCE}\neq\angle{PCD}$$ which is a contradiction.

The problem became when my friend asked me how can I say that $$CE=6$$ which caught me off guard. And now, after spending a few hours using trignometry to show that $$CE=6$$, I am clueless. The expressions are becoming too complicated.

So at the end I have two questions-

$$1)$$ How to show that $$CE=6$$?

$$2)$$ Is there any other way to show that the four will join up to a rhombus?

The problem is incorrect. The triangles need not make a rhombus, so that $$x$$ need not equal $$14$$.

Using Heron's formula, the triangles and their areas give us these equations: \begin{align} (a + b + 6)(-a + b + 6) (a - b + 6) (a + b - 6) &= 16\cdot4^2 \\ (b + c + 6)(-b + c + 6) (b - c + 6) (b + c - 6) &= 16\cdot5^2 \\ (c + d + 6)(-c + d + 6) (c - d + 6) (c + d - 6) &= 16\cdot x^2 \\ (d + a + 6)(-d + a + 6) (d - a + 6) (d + a - 6) &= 16\cdot13^2 \end{align} Each candidate value for area $$x$$ yields a solvable system in $$a$$, $$b$$, $$c$$, $$d$$. Using Mathematica to solve numerically gives these options:

$$\begin{array}{c|cccc|c} x & a & b & c & d & \text{angle sum} \\ \hline 12 & 12.50868 & \phantom{1}6.57405 & 12.47114 & 18.25586 & \phantom{9}25.17755^\circ \\ '' & \phantom{1}7.18734 & \phantom{1}1.70495 & \phantom{1}5.88180 & \phantom{1}4.34148 & 252.92081^\circ \\ \hline 13 & \phantom{1}7.56481 & \phantom{1}1.96718 & \phantom{1}7.23946 & \phantom{1}4.33798 & 185.40652^\circ \\ '' & \phantom{1}7.38591 & 13.33141 & \phantom{1}7.41651 & \phantom{1}4.33337 & 118.78946^\circ \\ \hline 14 & \phantom{1}4.53382 & \phantom{1}2.13437 & \phantom{1}4.95536 & \phantom{1}6.36832 & 360^\circ \\ \hline 15 & \phantom{1}4.74670 & \phantom{1}1.96573 & \phantom{1}5.23039 & \phantom{1}9.04321 & 301.05904^\circ \\ '' & \phantom{1}5.15927 & \phantom{1}1.67631 & \phantom{1}6.40040 & \phantom{1}5.38679 & 310.87315^\circ \end{array}$$ where "angle sum" is the total of the angles opposite the sides of length $$6$$.

The rhombus-ness in the $$x=14$$ case would seem to account for the single solution.

Another investigation to consider is eliminating, say, $$b$$, $$c$$, $$d$$ from the system, leaving a relation in $$x$$ and $$a$$. Mathematica makes light work of this, too. The result is a very ugly degree-$$8$$ polynomial in $$x^2$$ (which I won't enter here), indicating that $$a$$ can be varied continuously (within certain constraints) and yield valid (positive) roots $$x$$.

So, not only are there multiple possible answers, there's a continuum of such.

The problem is incorrect.

Let’s say that $$c=6$$ and see if this gives us a good counterexample. Given an acute triangle with two sides and an area, it’s possible to compute the third side (I just used an online calculator).

Since b,6,6 gives area 5, b can be computed as 1.6833.

Since a,1.6833,6 gives area 4, a can be 5.148.

Since d,5.148,6 gives area 13, d can be 5.399.

The area of c,d,6 would then be the area of a triangle with sides 6,5.399,6 which is 14.465.

Thus, it’s possible to satisfy all the constraints of the problem and have $$x$$ not be one of the choices.

I suppose the problem might still be possible if the constraints force some kind of range on the possible area which only contains 14, but I think a more likely solution is that the problem writers just screwed up.

• I think you are right. I was just trying a regression, and here it is. Feb 24 at 15:51
• I have taken the liberty of adding some diagrams to my post with a slightly modified form of your data. Is that all right? If it isn't, please say so. I'll rollback the changes. Feb 24 at 17:06

Why do you think ‘apparently… we get a rhombus’? It is not always true.

Any proof that assumes that the triangles join to form a rhombus is wrong, because we have a counter example: take $$a=b=c=d=6$$. Unless we use the areas somehow, we can never assure that they will join to form a rhombus.

The following was not my idea. It is merely a visualization of the data (slightly modified) given by @Eric in his post:

• I am quite confused. The idea that they join up to form a rhombus is not mine. I saw it in a youtube video. He didn't give any explanation in the video. Also this is just an mcq, then why would they make it so ambiguous. Feb 24 at 15:41
• I also think I saw the same proof somewhere on Youtube. MindYourDecisions maybe? Have you understood why the rhombus thing is wrong? Feb 24 at 15:46
• And I don't think the question is ambiguous. As far as I have experience with the ISI entrance questions, it's all about catching the right idea. Let's see if we can do it... Feb 24 at 15:49
• This is not a counterexample. If $a=b=c=d=6$, then the areas of the triangles are not 4,5,13 Feb 24 at 15:51
• I was not providing a counterexample to the question. I was merely asserting that without using the information about the area, we can't assume it's a rhombus. The solution that the OP found, involves assuming the rhombus first, which is clearly wrong. @Crostul Feb 24 at 15:52