Kvant Math Problem 95

Let the trapezoid have bases $AB$ and $CD$, with $AB>CD$, and let $E$ and $F$ be the midpoints of the legs.

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Verified: no
Verdicts: SKIP + SKIP
Solve time: 5m20s
Source on kvant.digital

Problem

A trapezoid was drawn on the board. Its midline $EF$ was drawn, and a perpendicular $OK$ was dropped from the point $O$ where the diagonals intersect to the longer base. Then the trapezoid was erased. How can the drawing be reconstructed from the remaining segments $EF$ and $OK$ (Fig. 2)?

Insert illustration here.

Exploration

Let the trapezoid have bases $AB$ and $CD$, with $AB>CD$, and let $E$ and $F$ be the midpoints of the legs. The segment $EF$ is the midline. The diagonals intersect at $O$, and $OK$ is perpendicular to the longer base $AB$.

The surviving data are only the segments $EF$ and $OK$. Since the trapezoid has been erased, the first question is whether these two segments determine the trapezoid uniquely.

Place coordinates so that the longer base is horizontal. Let

$$A=(0,0),\qquad B=b,\qquad D=(p,h),\qquad C=(p+c,h),$$

where $b>c>0$.

The midline joins the midpoints of the legs, hence

$$E=\left(\frac p2,\frac h2\right),\qquad F=\left(\frac{b+p+c}{2},\frac h2\right).$$

Therefore

$$EF=\frac{b+c}{2}.$$

The intersection point of the diagonals is found from the equations of $AC$ and $BD$. Solving gives

$$O=\left(\frac{b(p+c)}{b+c},\frac{bh}{b+c}\right).$$

Hence the distance from $O$ to the longer base equals

$$OK=\frac{bh}{b+c}.$$

Using $EF=(b+c)/2$, this becomes

$$OK=\frac{bh}{2,EF}.$$

The quantity $h$ is not yet known. A relation involving only the visible segments is needed.

The coordinates of $E$ and $F$ show that the line $EF$ lies halfway between the bases. Since the bases are horizontal, the distance from $EF$ to the longer base equals $h/2$. Thus, if $K$ is known, the longer base is the horizontal line through $K$, while $EF$ is already drawn. Consequently

$$h=2,d(EF,AB).$$

Since $K$ lies on $AB$, the distance from $K$ to the line $EF$ is exactly $h/2$. Therefore

$$h=2,d(K,EF).$$

Substituting into the previous formula yields

$$b=EF\cdot\frac{OK}{d(K,EF)}.$$

Thus the longer base length is completely determined by the given figure. Once $b$ is known,

$$c=2EF-b.$$

The ratio in which the diagonals divide one another in a trapezoid is

$$\frac{AO}{OC}=\frac{BO}{OD}=\frac{AB}{CD}=\frac bc.$$

Hence, after reconstructing the bases, the diagonals can be recovered from the point $O$ by reversing this ratio. The delicate point is to show that the positions of the endpoints of the bases are then uniquely determined.

Problem Understanding

We are given only two segments that remained after a trapezoid was erased: its midline $EF$ and the perpendicular $OK$ dropped from the intersection point $O$ of the diagonals to the longer base. The task is to reconstruct the original trapezoid using only these remaining segments.

This is a Type D problem. We must explicitly construct the trapezoid and verify that the construction indeed reproduces the original one.

The core difficulty is extracting the lengths of the two bases from the surviving data. The segment $EF$ gives only the sum of the bases, while $OK$ and the position of $K$ relative to $EF$ must be used to recover their individual lengths.

The answer is that the trapezoid is uniquely determined. The intuition is that $EF$ determines the average of the base lengths, while the location of $O$ relative to the bases encodes the ratio of the bases through the well-known division property of the diagonals.

Proof Architecture

Lemma 1. If the bases of a trapezoid have lengths $b$ and $c$, then its midline has length $(b+c)/2$.

Sketch. This is the standard property of the midline of a trapezoid.

Lemma 2. If $O$ is the intersection point of the diagonals, then the distance from $O$ to the longer base equals $bh/(b+c)$, where $h$ is the height.

Sketch. Compute the coordinates of $O$ in a coordinate model of the trapezoid.

Lemma 3. The height satisfies $h=2,d(K,EF)$.

Sketch. The midline lies halfway between the two bases.

Lemma 4. The longer base length is

$$b=EF\cdot\frac{OK}{d(K,EF)},$$

and the shorter base length is $c=2EF-b$.

Sketch. Combine Lemmas 1, 2, and 3.

Lemma 5. The diagonals satisfy

$$\frac{AO}{OC}=\frac{BO}{OD}=\frac bc.$$

Sketch. This is the standard proportionality theorem for diagonals of a trapezoid.

The most delicate point is Lemma 4, because it extracts the actual base lengths from the remaining segments.

Solution

Let the unknown trapezoid be $ABCD$, where $AB$ is the longer base and $CD$ is the shorter base. Let $E$ and $F$ be the midpoints of the legs, and let $O$ be the intersection point of the diagonals. The given figure consists only of the segments $EF$ and $OK$, where $OK\perp AB$.

Denote

$$m=EF,$$

and let

$$\delta=d(K,EF).$$

Since $EF$ is the midline of the trapezoid, it is parallel to the bases and lies halfway between them. Therefore the height $h$ of the trapezoid satisfies

$$h=2\delta.$$

Let the base lengths be

$$AB=b,\qquad CD=c,$$

with $b>c$.

By the midline theorem,

$$m=\frac{b+c}{2}.$$

Choose coordinates

$$A=(0,0),\quad B=(b,0),\quad D=(p,h),\quad C=(p+c,h).$$

A direct computation of the intersection point of the diagonals gives

$$O=\left(\frac{b(p+c)}{b+c},\frac{bh}{b+c}\right).$$

Hence the distance from $O$ to the longer base is

$$OK=\frac{bh}{b+c}.$$

Using $b+c=2m$ and $h=2\delta$, we obtain

$$OK=\frac{b(2\delta)}{2m} =\frac{b\delta}{m}.$$

Therefore

$$b=m\frac{OK}{\delta}.$$

Since $b+c=2m$,

$$c=2m-b.$$

Thus both base lengths are determined uniquely from the given segments.

Now reconstruct the trapezoid.

Draw the line through $K$ parallel to $EF$. This is the longer base line $AB$. Reflect it across the line containing $EF$; the image is the shorter base line $CD$, because the midline is halfway between the bases.

On the line through $K$, mark a segment $AB$ of length

$$b=m\frac{OK}{\delta},$$

so that the foot of the perpendicular from $O$ to the line is the given point $K$.

On the parallel line $CD$, mark a segment of length

$$c=2m-b.$$

Since in every trapezoid

$$\frac{AO}{OC}=\frac{BO}{OD}=\frac bc,$$

draw through $O$ a line meeting the two base lines at points $A$ and $C$ so that

$$\frac{AO}{OC}=\frac bc.$$

Draw through $O$ another line meeting the base lines at points $B$ and $D$ so that

$$\frac{BO}{OD}=\frac bc.$$

The four obtained points determine a trapezoid. Its bases have lengths $b$ and $c$, its midline is the given segment $EF$, and the perpendicular from the diagonal intersection point to the longer base is the given segment $OK$.

Hence the original trapezoid is reconstructed.

$$\boxed{\text{The trapezoid obtained by the construction above}}$$

Verification of Key Steps

The first delicate step is the formula

$$OK=\frac{bh}{b+c}.$$

Using coordinates,

$$AC:\ (0,0)+t(p+c,h),$$

and

$$BD:\ (b,0)+s(p-b,h).$$

At the intersection, the $y$-coordinates are equal, hence $t=s$. Equating the $x$-coordinates gives

$$t(p+c)=b+t(p-b),$$

so

$$t(b+c)=b, \qquad t=\frac b{b+c}.$$

The $y$-coordinate of $O$ is therefore

$$th=\frac{bh}{b+c},$$

which is exactly $OK$.

The second delicate step is the recovery of the height. Since the midline joins the midpoints of the legs, its points have height $h/2$ above the longer base and height $h/2$ below the shorter base. Thus the distance from either base to the midline equals $h/2$. Because $K$ lies on the longer base,

$$\delta=d(K,EF)=\frac h2,$$

and therefore $h=2\delta$.

The third delicate step is obtaining the base lengths. Substituting $h=2\delta$ and $b+c=2m$ into

$$OK=\frac{bh}{b+c}$$

gives

$$OK=\frac{b\delta}{m},$$

hence

$$b=m\frac{OK}{\delta}.$$

No additional unknown remains, so both bases are uniquely determined.

Alternative Approaches

A synthetic solution avoids coordinates. Let $P$ be the point where the diagonal $AC$ meets the midline. Since the midline lies halfway between the bases, $P$ is the midpoint of $AC$. The same holds for the second diagonal. Using the fact that the diagonals of a trapezoid divide one another in the ratio of the bases, one obtains

$$\frac{AO}{OC}=\frac{AB}{CD}.$$

The distances from $O$ and the midpoint of a diagonal to the base are proportional to the same ratio, which yields the relation between $OK$, the height, and the base lengths. From there the reconstruction proceeds as above.

The coordinate method is preferable because it produces directly the explicit formula

$$b=EF\cdot\frac{OK}{d(K,EF)},$$

from which the construction follows immediately.