Why would the triangles join up to a rhombus?

Question

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?

Answer 1

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.

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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.

Answer 2

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.

Answer 3

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: