Kvant Math Problem 183
Let the trapezoid have bases of lengths $b$ and $a$, with $a<b$.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 5m41s
Source on kvant.digital
Problem
Find the height of a trapezoid whose bases are $a$ and $b$ ($a \lt b$), the angle between its diagonals is $90^{\circ}$, and the angle between the extensions of its lateral sides is $45^{\circ}$.
Exploration
Let the trapezoid have bases of lengths $b$ and $a$, with $a<b$. Put the longer base on the $x$-axis. Let the lower base be $AB$ with $A=(0,0)$ and $B=(b,0)$. Let the upper base be $CD$ with $D=(p,h)$ and $C=(p+a,h)$.
The diagonals are
$$AC=(p+a,h),\qquad BD=(p-b,h).$$
Since they are perpendicular,
$$(p+a)(p-b)+h^2=0.$$
Hence
$$h^2=(b-p)(p+a).$$
The extensions of the lateral sides $AD$ and $BC$ meet at a point $E$. Because $AB\parallel CD$, triangles $EAB$ and $EDC$ are similar. If $t=\dfrac{ED}{EA}$, then
$$t=\frac{DC}{AB}=\frac ab.$$
Along the side $EA$,
$$D=A+t(E-A).$$
Since $A=(0,0)$ and $D=(p,h)$,
$$E=\left(\frac{bp}{a},\frac{bh}{a}\right).$$
The angle between the extensions of the lateral sides is $\angle DEB=45^\circ$. Using vectors
$$\overrightarrow{ED}=D-E,\qquad \overrightarrow{EC}=C-E,$$
their slopes are proportional to
$$(a-b)h,\qquad ap-b(p+a).$$
A cleaner computation is obtained from
$$\overrightarrow{ED}=\frac{b-a}{a}(-p,-h),\qquad \overrightarrow{EC}=\frac{b-a}{a}(b-p-a,-h).$$
The factor $\frac{b-a}{a}$ is irrelevant.
Thus the angle condition becomes
$$\frac{|(-p,-h)\cdot(b-p-a,-h)|} {\sqrt{p^2+h^2}\sqrt{(b-p-a)^2+h^2}} =\frac{\sqrt2}{2}.$$
The algebra looks unpleasant. Introduce
$$x=p,\qquad y=b-a-p.$$
Then
$$h^2=(b-p)(p+a)=(a+x)(y+a),$$
because $x+y=b-a$.
The $45^\circ$ condition gives
$$(xy-h^2)^2=(x^2+h^2)(y^2+h^2).$$
Expanding and cancelling yields
$$h^2(x+y)^2=2(xy-h^2)^2.$$
Since $x+y=b-a$ and
$$xy-h^2=xy-(a+x)(a+y)=-a(a+x+y)=-ab,$$
we obtain
$$h^2(b-a)^2=2a^2b^2.$$
Therefore
$$h=\frac{\sqrt2,ab}{b-a}.$$
The crucial point is the identity $xy-h^2=-ab$.
Problem Understanding
We are given a trapezoid with bases $a$ and $b$, where $a<b$. Its diagonals are perpendicular, and the angle formed by the extensions of the two lateral sides is $45^\circ$. The task is to determine the height of the trapezoid.
This is a Type C problem, since a specific numerical quantity must be found from the given geometric conditions.
The core difficulty is translating the two angle conditions into algebraic relations and eliminating the position of the upper base. The answer should depend only on $a$ and $b$, because the conditions determine the trapezoid uniquely up to similarity.
The expected answer is
$$h=\frac{\sqrt2,ab}{b-a}.$$
Proof Architecture
Introduce coordinates with the bases horizontal and denote the horizontal displacement of the upper base by $p$.
Use the perpendicularity of the diagonals to obtain
$$h^2=(b-p)(p+a).$$
Find the intersection point of the extensions of the lateral sides by similarity of the triangles cut out by the parallel bases.
Express the vectors along the two lateral sides through this intersection point and impose the condition that the angle between them is $45^\circ$.
Set
$$x=p,\qquad y=b-a-p,$$
so that $x+y=b-a$ and
$$h^2=(a+x)(a+y).$$
Convert the $45^\circ$ condition into
$$h^2(x+y)^2=2(xy-h^2)^2.$$
Prove the identity
$$xy-h^2=-ab.$$
Substitute the last two relations and solve for $h$.
The most delicate step is deriving the equation
$$h^2(x+y)^2=2(xy-h^2)^2$$
from the $45^\circ$ condition and simplifying it correctly.
Solution
Let the trapezoid be $ABCD$, where $AB=b$, $CD=a$, and $AB\parallel CD$.
Choose coordinates
$$A=(0,0),\qquad B=(b,0),$$
and let
$$D=(p,h),\qquad C=(p+a,h).$$
The diagonals are represented by the vectors
$$\overrightarrow{AC}=(p+a,h),\qquad \overrightarrow{BD}=(p-b,h).$$
Since the diagonals are perpendicular,
$$\overrightarrow{AC}\cdot\overrightarrow{BD}=0,$$
hence
$$(p+a)(p-b)+h^2=0.$$
Therefore
$$h^2=(b-p)(p+a). \tag{1}$$
Let $E$ be the intersection point of the extensions of the lateral sides $AD$ and $BC$.
Because $AB\parallel CD$, triangles $EAB$ and $EDC$ are similar. Therefore
$$\frac{ED}{EA}=\frac{DC}{AB}=\frac ab.$$
Since $D$ lies on segment $EA$,
$$D=A+\frac ab(E-A).$$
With $A=(0,0)$ and $D=(p,h)$ this gives
$$E=\left(\frac{bp}{a},\frac{bh}{a}\right).$$
Hence
$$\overrightarrow{ED} =\left(p-\frac{bp}{a},,h-\frac{bh}{a}\right) =\frac{b-a}{a}(-p,-h),$$
and
$$\overrightarrow{EC} =\left(p+a-\frac{bp}{a},,h-\frac{bh}{a}\right) =\frac{b-a}{a}(b-p-a,-h).$$
Multiplying vectors by the same nonzero factor does not change the angle between them. Thus we may work with
$$u=(-p,-h),\qquad v=(b-p-a,-h).$$
Introduce
$$x=p,\qquad y=b-a-p.$$
Then
$$u=(-x,-h),\qquad v=(y,-h),$$
and
$$x+y=b-a. \tag{2}$$
The angle between $u$ and $v$ equals $45^\circ$. Using
$$\cos45^\circ=\frac{\sqrt2}{2},$$
we obtain
$$\frac{|u\cdot v|} {\sqrt{u\cdot u}\sqrt{v\cdot v}} =\frac{\sqrt2}{2}.$$
After squaring,
$$2(xy-h^2)^2=(x^2+h^2)(y^2+h^2). \tag{3}$$
Expanding the right-hand side and subtracting the left-hand side,
$$0=x^2y^2+h^2(x^2+y^2)+h^4 -2x^2y^2+4xyh^2-2h^4.$$
Thus
$$0=-(xy-h^2)^2+h^2(x+y)^2,$$
which is equivalent to
$$h^2(x+y)^2=2(xy-h^2)^2. \tag{4}$$
From (1),
$$h^2=(b-p)(p+a).$$
Using the definitions of $x$ and $y$,
$$b-p=a+y,$$
so
$$h^2=(a+x)(a+y). \tag{5}$$
Consequently,
$$xy-h^2 =xy-(a+x)(a+y) =-a(a+x+y).$$
By (2),
$$a+x+y=a+(b-a)=b.$$
Therefore
$$xy-h^2=-ab. \tag{6}$$
Substituting (2) and (6) into (4),
$$h^2(b-a)^2=2a^2b^2.$$
Since $h>0$,
$$h=\frac{\sqrt2,ab}{b-a}.$$
Hence the height of the trapezoid is
$$\boxed{\frac{\sqrt2,ab}{,b-a,}}.$$
Verification of Key Steps
The first delicate point is locating the intersection point $E$ of the extensions of the lateral sides. Similarity of triangles $EAB$ and $EDC$ yields
$$\frac{ED}{EA}=\frac ab.$$
Since $D$ divides segment $EA$ in the ratio $a:b$, the coordinate relation
$$D=A+\frac ab(E-A)$$
follows directly. Solving for $E$ gives
$$E=\left(\frac{bp}{a},\frac{bh}{a}\right),$$
which agrees with both coordinates of $D$.
The second delicate point is the conversion of the $45^\circ$ condition. Starting from
$$\frac{|u\cdot v|}{|u||v|}=\frac{\sqrt2}{2},$$
squaring gives
$$2(u\cdot v)^2=|u|^2|v|^2.$$
For
$$u=(-x,-h),\qquad v=(y,-h),$$
this becomes equation (3). Any omission of the squaring step or of the factor $2$ leads to an incorrect final height.
The third delicate point is the identity
$$xy-h^2=-ab.$$
Using
$$h^2=(a+x)(a+y)$$
and
$$x+y=b-a,$$
one obtains
$$xy-h^2=-a(a+x+y)=-ab.$$
This cancellation is the mechanism that removes the parameter $p$ and leaves an answer depending only on the bases.
Alternative Approaches
A synthetic solution can be built around the point where the extensions of the lateral sides meet. Let this point be $E$. Because $AB\parallel CD$, the homothety centered at $E$ sending $AB$ to $CD$ has ratio $a/b$. Distances from $E$ to the two bases can then be expressed through the height of the trapezoid. The condition that the angle between the sides through $E$ is $45^\circ$ converts these distances into relations involving tangents of half-angles.
The perpendicularity of the diagonals can also be interpreted through powers of the homothety center and similar triangles generated by the diagonals. After eliminating the auxiliary lengths introduced by the homothety, one again arrives at
$$h^2(b-a)^2=2a^2b^2.$$
The coordinate method is preferable because both geometric conditions translate directly into quadratic equations, and the elimination of the displacement parameter is completely transparent.