# Slicing a rectangle

A friend presented this nice little puzzle to me yesterday. You're given a rectangle which is dissected by one of its diagonals, as well as another line that only meets one of the two remaining corners.

The smallest triangle's area is $2$ and the adjacent triangle's area is $3$. What is the area of the quadrilateral?

You are not given the aspect ratio of the rectangle, nor the angle of the line separating the $2$ from the $?$ (nor the length ratios of any line segments). The above diagram is not necessarily to scale.

The area of ? is:

$\frac{11}{2}$

Because:

Name the triangles and angles as such:
Now the $\angle\alpha$ are the same because they're opposite angles.
And the $\angle\beta$ are the same because they're alternate angles.
Likewise $\angle\theta$ are the same because they're also alternate angles.
So $A$ is similar to $C$ since the angles are the same.

Working from there:

Consider the following heights and segments of these triangles:
Now because $A$ and $C$ are similar $\frac{c}{h} = \frac{l}{a-l}$
So $c = \frac{h l}{a-l} \tag1$
Area of $A = \frac12 l (c + h) \tag2$
Plugging $(1)$ into $(2)$ yields:
$\frac12 l \left(\frac{h l}{a-l} + h \right) = \frac12 l \left( \frac{h l + h a - h l}{a - l} \right) = \frac12 \frac{a h l}{a-l} = A \tag3$
Now area of $B = \frac12 a h \tag4$
Plugging $(4)$ into $(3)$ gives us $\frac{B l}{a-l} = A\tag5$
Now the ratio of sides of the similar triangles $A$ and $C$ is the same as the ratio of their heights and from $(5)$ we get $\frac{l}{a-l} = \frac{A}{B} = r$.
Again because they're similar, this gives the area of $C = r^{-2}\cdot A = \left({{\frac{B}{A}}}\right)^2 \cdot A = \frac{B^2}{A}$.
Since the main diagonal slices the area of the rectangle in half we have areas ? $+ A = B + C$
And from that we get the area of ? $= B + C - A = B + \frac{B^2}{A} - A = \frac{A B + B^2 - A^2}{A} = \frac{11}{2}$.

First of all, by a simple geometry principle:

• $\triangle CED$ and $\triangle AED$ have the same base $|ED|$ and the area ratio between $\triangle CEF$ and $\triangle CFD$ has to be the same as the length ratio between $|EF|$ and $|FD|$ so, similarly, you can tell the area ratio between $\triangle AEF$ and $\triangle AFD$ as $2x$ and $3x$.

Notice from this image that $\triangle AED$ and $\triangle ACD$ are both half the area of the rectangle.

$2x + 3x = 3 + 3x$
$2x = 3$
$x = 1.5$

And then

$R = 2(2x + 3x)$
$R = 10x$
$R = 15$ — the area of the rectangle is 15.

and

$R - 5 - 3x = Q$
$10 - 3x = Q$
$5.5 = Q$ — the area of the quadrilateral is 5.5.

• What's the simple geometry principle? – Narmer Mar 31 '16 at 13:16
• @Narmer, if you want, I can explain every single detail but I assumed everyone knows the basic principles of a geometry for same base and same height with the same area... – Oray Mar 31 '16 at 13:20
• It must be in plain sight, but I can't see the same height and same base for any two triangles... – Narmer Mar 31 '16 at 13:21
• I mean, I can see the triangles created by the diagonal, I can't just see how you deduced $3x$ and $2x$ from it – Narmer Mar 31 '16 at 13:26
• I'm also confused on how you deduced So we can easily say that if $|CF|=1$ while calculating $7x-5$. How are them related? Why $1$? – Narmer Mar 31 '16 at 13:31

Now that we have two correct answers, I figured I'd present my own approach. It's similar to Paul's but doesn't work with the ratios of the side lengths, but instead directly with the ratios of the areas.

I'm splitting the solution into several spoiler paragraphs so that they can be used as individual hints towards the solution.

The first observation is that the problem is actually independent of the aspect ratio of the rectangle: if the entire diagram is stretched along one of the axes, the area ratios will remain unchanged, because all areas are scaled equally.

With that in mind, we can extend the line that separates $?$ and $2$ to form a new rectangle:

We can now see quite easily that in the following diagrams the ratio between the red and the blue segment is identical (consider the projection of the diagonal onto the other two lines):

That means, up to rotating by 90 degrees and some stretching, the new rectangle has the same setup as the original one, but with the roles of the two lines swapped:

So the ratio of the area of the triangle that was originally labelled $3$ and the unlabelled triangle is also $\frac{3}{2}$, so we get an area of $\frac{9}{2}$ for the unlabelled triangle:

Now it's just a matter of noting that the diagonal splits the rectangle into two equal triangles, so we can equate their areas.

\begin{align} ? + 2 &= \frac{9}{2} + 3 \\ ? &= \frac{11}{2} \end{align}

As a corollary, note that none of this depends on the given ratio being specifically $\frac{3}{2}$. So we can easily generalise this to a ration $\frac{B}{A}$ between the smallest and its adjacent triangle:

What we've really shown is that the three triangles are always in a geometric progression.

The general solution is then $$? = \frac{B^2 + AB - A^2}{A}$$

Here's an approach that I think is easier than the other approaches...

Let the area of the unknown triangle and quadrilateral be T and Q respectively.

Note that T and the triangle of area 2 are similar because all of their angles are the same. Their sides are in the ratio 3 to 2 since that is the ratio of the area of the two given triangles.

This means their areas are in the ratio $(\frac{3}{2})^2$, so $T=2 \times (\frac{3}{2})^2 = \frac{9}{2}$

Finally, $2+Q=3+T$, so $Q=\frac{11}{2}$.

• This is what Paul Evans's answer said before he edited it. – f'' Mar 31 '16 at 23:25
• I did not realize that. It seems he made his answer significantly more complicated, so I will keep this up as a simpler approach that leaves a few details to the reader. – DaveBlackston Mar 31 '16 at 23:29
• I was getting comments about "how do you know A and C similar", etc. So I thought I'd better make every thing crystal clear. That does have the bonus of getting $\frac{B^2+BA-A^2}{A}$ as the solution. But I do like this solution (my original) better :) – Paul Evans Apr 1 '16 at 18:48

Let the vertices of the rectangle be $(0,0), (a,0), (0,b)$, and $(a,b)$.

The positive-slope diagonal line is $y = (\frac{b}{a})x$. The negative-slope line meets the boundary of the rectangle at vertex $(a,0)$, and also at some point $(c,b)$. The equation of this line is $y = \left(\frac{b}{c-a} \right)(x-a)$.

We calculate that the intersection point of the two lines is $(x,y) = \left( \frac{a^2}{2a-c}, \frac{ab}{2a-c} \right)$.

Now, the area of the smaller triangle is $2$, so

\begin{eqnarray}2& =& \left(\frac{1}{2}\right)(a-c)(b-y)\\4(c-2a) &=& (a-c)b(c-2a) + (a-c)ab\phantom{***}(*)\end{eqnarray}

On the other hand, consider the triangle with area $2+3 = 5$:

$5 = \left( \frac{1}{2} \right) b(a-c)$,

so

$b = 10/(a-c)$.

Substituting into equation $(*)$, we get

\begin{eqnarray*}4(c-2a) &=& 10(c-2a) + 10a\\a &=& 3c\end{eqnarray*}

and therefore

$b = 5/c$.

At this point we see that

$Area(rectangle) = ab = (3c)(5/c) = 15$

Thus

\begin{eqnarray*}Area(?) &= &15 - 2 - 3 - \left( \frac{1}{2} \right)ay\\&=& 10 - \left( \frac{1}{2} \right) \left(\frac{a^2b}{2a-c}\right)\\&=& 10 - \left( \frac{1}{2} \right) \left(\frac{(9c^2)(5/c)}{6c-c}\right)\\&=& 10 - \frac{9}{2}\\& =& \frac{11}{2}.\end{eqnarray*}