# Minimum area of whole quadrilateral given areas of parts

I've attempted to mark segment $$AO$$ as $$a$$ and $$CO$$ as $$b$$.

Now, if draw a line from $$D$$ that is perpendicular to $$AC$$, and draw another line from B that is perpendicular to $$AC$$, then we can use some variables to express the area of $$\triangle BOC$$ and $$\triangle AOD$$.

However, when I try to make a equation, I found that I have too many variables.

How can I move further?

• @cosmo5 That means we should find a solution based on junior high school knowledge. No calculus and not too deep into trigonometric functions, etc. Nov 24, 2020 at 13:12

We need to use the fact that the ratio of triangles with the same height is equal to the ratio of their bases.

Suppose $$AO : OC = 1: k$$. Then $$[\triangle AOD] : [\triangle COD] = AO:OC = [\triangle AOB] : [\triangle COB] = 1:k$$.

This gives $$[\triangle AOD] = \frac4k$$ and $$[\triangle COB] = 9k$$.

Now the area of the quadrilateral is equal to

$$4 + 9+\frac4k+9k=4+9+12+\frac4k-12+9k=25+\left(\frac{2}{\sqrt k}-3\sqrt{k}\right)^2 \ge 25$$

Equality holds when $$k = \frac23$$, so the minimum area is $$25$$.

We can use $$AM \ge GM: a+b \ge 2\sqrt {ab}$$ to obtain $$\dfrac 4k + 9k \ge 12$$ as well.

Altitude from $$B$$ to $$AC$$, $$h_1$$ is $$\dfrac{2[AOB]}{AO}=\dfrac{2\triangle_{1}}{a}$$

Altitude from $$D$$ to $$AC$$, $$h_2$$ is $$\dfrac{2[COD]}{CO}=\dfrac{2\triangle_{2}}{b}$$

So $$[ABCD] = [AOB]+[COD]+[BOC]+[DOA]$$

$$= \triangle_{1}+\triangle_{2}+\dfrac{1}{2}\cdot b \cdot \dfrac{2\triangle_{1}}{a}+\dfrac{1}{2}\cdot a \cdot\dfrac{2\triangle_{2}}{b}$$

$$= \triangle_{1}+\triangle_{2}+ \dfrac{b}{a}\triangle_{1}+ \dfrac{a}{b}\triangle_{2}$$

which is minimum when (their product is constant so their sum is minimum when both equal)

$$\dfrac{b}{a}\triangle_{1} = \dfrac{a}{b}\triangle_{2}$$ $$\Rightarrow \dfrac{\triangle_{1}}{\triangle_{2}} = \dfrac{a^2}{b^2}$$

Obtain same expression by taking altitudes on other diagonal.

Since $$\angle AOB = \angle COD$$, we conclude $$[ABCD]$$ is minimum when $$\triangle AOB \sim \triangle COD$$

As a result $$\angle BAO = \angle DCO \Rightarrow AB || CD$$

Thus $$ABCD$$ is a trapezium.

You can substitute back to find the minimum value is

$$\bbox[#FFFF33,5px]{[ABCD]_{\text{min}} = (\sqrt{\triangle_{1}}+\sqrt{\triangle_{2}})^2}$$