#6 Areas I

By the area formula, \(S=\frac{1}{2}\cdot18\cdot7=63\).

Both triangles have the height from point \(A\) to line \(BC\). Therefore their areas are in the ratio of their bases on this line: \(S_{ABD}:S_{ACD}=BD:DC\).

Since \(M\) is the midpoint of \(BC\), \(BM=CM\). Triangles \(ABM\) and \(ACM\) have the common height from \(A\) to line \(BC\). Therefore their areas are equal.

Triangles \(ABC\) and \(ABD\) have common base \(AB\). The heights from \(C\) and \(D\) to \(AB\) are equal because these points lie on a line parallel to \(AB\). Hence the areas are equal.

Let \(ABCD\) be a parallelogram. Diagonal \(AC\) divides it into triangles \(ABC\) and \(ACD\). They have common base \(AC\), and the heights from \(B\) and \(D\) to \(AC\) are equal because \(AB\parallel CD\). Hence these triangle areas are equal.

Since \(BD:DC=3:4\), we have \(BD:BC=3:7\). Triangles \(ABD\) and \(ABC\) have the common height from \(A\) to \(BC\). Therefore \(S_{ABD}:S_{ABC}=BD:BC=3:7\).

Since \(CD\parallel AB\), points \(C\) and \(D\) lie on one line parallel to base \(AB\). Therefore the heights to base \(AB\) are equal. The base is common, so the areas are equal.

Median \(AM\) divides the triangle into two equal-area triangles. Therefore \(S_{ABM}=\frac{84}{2}=42\).

We have \(BD=\frac{1}{2}BA\) and \(BE=\frac{1}{2}BC\), and the angle at \(B\) is common. The area is reduced by \(2\cdot2=4\) times. Therefore \(S_{BDE}=\frac{64}{4}=16\).

Triangles \(ABC\) and \(ACD\) have bases \(BC\) and \(AD\) on parallel lines, and the heights to these bases equal the distance between the bases of the trapezoid. Therefore \(S_{ABC}:S_{ACD}=BC:AD=9:15=3:5\).

Triangles \(ABD\) and \(ACD\) have the common height from \(A\). Therefore \(BD:DC=S_{ABD}:S_{ACD}=24:40=3:5\).

Since base \(AB\) is common and the areas are equal, the heights from \(C\) and \(D\) to \(AB\) are equal. Points on the same side of \(AB\) at the same distance from \(AB\) lie on a line parallel to \(AB\). Hence \(CD\parallel AB\).

Denote the areas of \(AGF\), \(BGF\), \(BGD\), \(CGD\), \(CGE\), \(AGE\) in order by \(x_1,x_2,x_3,x_4,x_5,x_6\). Since \(F,D,E\) are side midpoints, we get \(x_1=x_2\), \(x_3=x_4\), \(x_5=x_6\). Median \(AD\) halves the whole triangle, so \(x_1+x_2+x_3=x_4+x_5+x_6\). Median \(BE\) gives \(x_1+x_2+x_6=x_3+x_4+x_5\). Substituting the pairwise equalities, we get \(x_1=x_5\) and \(x_1=x_3\). Hence all six areas are equal.

Since \(DE\parallel BC\), \(\triangle ADE\sim\triangle ABC\). We have \(AD:AB=2:5\). Therefore the area ratio equals the square of the side ratio: \(S_{ADE}:S_{ABC}=4:25\).

Triangles \(ABD\), \(ADE\), \(AEC\) have bases \(BD\), \(DE\), \(EC\) on one line \(BC\) and a common height from \(A\). Since the bases are equal, the areas are equal.

Let \(AP:PC=t:(1-t)\). In triangle \(ABC\), \(S_{ABP}:S_{PBC}=AP:PC\), so \(S_{ABP}=tS_{ABC}\). Similarly, in triangle \(ACD\), \(S_{ADP}=tS_{ACD}\). But a diagonal of a parallelogram divides it into two equal-area triangles: \(S_{ABC}=S_{ACD}\). Hence \(S_{ABP}=S_{ADP}\).

Since \(DE\parallel AC\), \(\triangle BDE\sim\triangle BCA\). We have \(BD:BC=2:5\). Therefore areas are in the ratio of squares of corresponding sides: \(S_{BDE}:S_{ABC}=4:25\).

Since \(P\) lies on \(AM\), line \(AP\) is the median and passes through the midpoint \(M\) of side \(BC\). Triangles \(PAB\) and \(PAC\) have common base \(AP\). Points \(B\) and \(C\) are at equal distances from line \(AP\), so the areas are equal.

Triangles \(PAB\) and \(PAC\) have common base \(AP\). Equality of their areas implies that the distances from \(B\) and \(C\) to line \(AP\) are equal. Since \(B\) and \(C\) lie on opposite sides of \(AP\), line \(AP\) passes through the midpoint of \(BC\). Thus \(AP\) is a median.

Triangles \(BCE\) and \(BCA\) have common base \(BC\). Since \(AE:ED=1:2\), point \(E\) is \(\frac{1}{3}\) of the way from \(A\) to \(D\), where \(D\) lies on \(BC\). Therefore the height from \(E\) to \(BC\) is \(\frac{2}{3}\) of the height from \(A\). Hence \(S_{BCE}=\frac{2}{3}\cdot90=60\).

From \(S_{PAB}=S_{PCA}\), line \(AP\) passes through the midpoint of \(BC\), so \(AP\) is a median. From \(S_{PAB}=S_{PBC}\), line \(BP\) passes through the midpoint of \(AC\), so \(BP\) is a median. Therefore \(P\) is the intersection of two medians, hence the centroid of the triangle.

Triangles \(BCE\) and \(BCA\) have common base \(BC\). The height of point \(E\) to \(BC\) is \(ED:AD\) of the height of point \(A\), because \(D\) lies on \(BC\). From \(AE:ED=3:2\), we get \(ED:AD=2:5\). Therefore \(S_{BCE}=\frac{2}{5}\cdot120=48\).

Let the angle between the diagonals be \(\varphi\). Then \(S_{AOB}=\frac{1}{2}AO\cdot BO\sin\varphi\), \(S_{COD}=\frac{1}{2}CO\cdot DO\sin\varphi\). Similarly, \(S_{BOC}=\frac{1}{2}BO\cdot CO\sin\varphi\), \(S_{DOA}=\frac{1}{2}DO\cdot AO\sin\varphi\). The products on the left and right are identical, so the equality is proved.

Triangles \(ABD\) and \(ACD\) have common base \(AD\). The heights from \(B\) and \(C\) to \(AD\) are equal because \(BC\parallel AD\). Therefore \(S_{ABD}=S_{ACD}\). Subtract the common triangle \(AOD\) from both areas: we get \(S_{AOB}=S_{COD}\).

Triangles \(BDE\) and \(BCA\) are similar because \(DE\parallel AC\). The similarity ratio is \(BD:BC=2:5\). Therefore \(S_{BDE}:S_{ABC}=4:25\). The remaining part \(ADEC\) has area \(21\) parts out of \(25\). Hence \(S_{BDE}:S_{ADEC}=4:21\).

Let the distances from \(P\) to lines \(AB\) and \(CD\) be \(h_1\) and \(h_2\). Then \(h_1+h_2\) equals the height of the parallelogram to base \(AB\). Therefore \(S_{PAB}+S_{PCD}=\frac{1}{2}AB\cdot h_1+\frac{1}{2}CD\cdot h_2\). Since \(AB=CD\), this is \(\frac{1}{2}AB(h_1+h_2)=\frac{1}{2}S_{ABCD}\).

Since \(P,A,D\) are collinear, triangles \(ABP\) and \(PBD\) have heights to this line from \(B\), and triangles \(ACP\) and \(PCD\) have heights from \(C\). Comparing fractions on the same line gives \(\frac{BD}{DC}=\frac{S_{ABP}}{S_{ACP}}\). Similarly, \(\frac{CE}{EA}=\frac{S_{BCP}}{S_{ABP}}\) and \(\frac{AF}{FB}=\frac{S_{ACP}}{S_{BCP}}\). Multiplying, we get \(1\).

By the area form of Ceva's theorem, \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1\). Substitute: \(\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{AF}{FB}=1\). We get \(\frac{1}{2}\cdot\frac{AF}{FB}=1\), hence \(AF:FB=2:1\).

Since \(G\) lies on the median from \(A\), we have \(S_{GAB}=S_{GAC}\). Since \(G\) lies on the median from \(B\), we have \(S_{GAB}=S_{GBC}\). Hence \(S_{GAC}=S_{GBC}\). Triangles \(GAC\) and \(GBC\) have common base \(GC\), so points \(A\) and \(B\) are at equal distances from line \(GC\). Therefore line \(GC\) passes through the midpoint of \(AB\), that is, \(AF=FB\).

We have \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=\frac{2}{3}\cdot\frac{3}{5}\cdot\frac{5}{2}=1\). By the converse form of the area version of Ceva's theorem, product \(1\) implies that lines \(AD\), \(BE\), \(CF\) are concurrent, meaning they meet at one point.

E. Full solution.

Assume such a square has been tiled, and let \(1\times n\) be the longest chosen rectangle. Since the square has area greater than \(1\), we have \(n>1\).

The rectangle \(1\times n\) must fit inside the square, even if rotated. Therefore the side length of the square is at least \(n\), and its area is at least \(n^2\).

On the other hand, the chosen rectangles can only be among \(1\times1,1\times2,\ldots,1\times n\). Their total area is at most \(1+2+\cdots+n=\frac{n(n+1)}{2}\).

For \(n>1\), \(\frac{n(n+1)}{2}

Therefore a square of area greater than \(1\) cannot be tiled in this way.