Kvant Math Problem 139

Let the parallelogram have side vectors $\mathbf u=\overrightarrow{BA}$ and $\mathbf v=\overrightarrow{BC}$.

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Problem

From vertex $B$ of the parallelogram $ABCD$, its altitudes $BK$ and $BH$ are drawn. The segments $KH=a$ and $BD=b$ are known. Find the distance from point $B$ to the intersection point of the altitudes of triangle $BKH$.

F. A. Bartenev

Exploration

Let the parallelogram have side vectors $\mathbf u=\overrightarrow{BA}$ and $\mathbf v=\overrightarrow{BC}$. Since $BK$ and $BH$ are altitudes from $B$, point $K$ is the projection of $B$ onto line $CD$, while $H$ is the projection of $B$ onto line $AD$.

Because $CD\parallel AB$ and $AD\parallel BC$, the feet are projections onto lines parallel to $\mathbf u$ and $\mathbf v$. Writing everything in vector form should reveal the geometry of triangle $BKH$.

Choose coordinates with $B=(0,0)$, $\mathbf u=(m,0)$, $\mathbf v=(p,q)$. Then $A=(m,0)$, $C=(p,q)$, $D=(m+p,q)$.

The line $AD$ has direction $\mathbf v$, hence

$$H=A-\frac{A\cdot \mathbf v}{|\mathbf v|^2}\mathbf v.$$

The line $CD$ has direction $\mathbf u$, hence

$$K=C-\frac{C\cdot \mathbf u}{|\mathbf u|^2}\mathbf u.$$

A direct computation gives

$$K=(0,q),$$

and

$$H=\left(m\frac{q^2}{p^2+q^2},-\frac{mpq}{p^2+q^2}\right).$$

Now compute

$$K\cdot H = 0\cdot x_H+q\left(-\frac{mpq}{p^2+q^2}\right) = -\frac{mpq^2}{p^2+q^2}.$$

Also

$$|K|^2=q^2.$$

Hence

$$(K-H)\cdot K = |K|^2-K\cdot H = q^2+\frac{mpq^2}{p^2+q^2}.$$

A calculation of $|K-H|^2$ yields

$$|K-H|^2 = q^2\left(1+\frac{mp}{p^2+q^2}\right)^2.$$

Therefore

$$(K-H)\cdot K=|K-H|^2.$$

Since

$$(K-H)\cdot(K-(K-H))=0,$$

we obtain

$$BH\perp HK.$$

Thus triangle $BKH$ is right-angled at $H$.

This is the crucial observation. In a right triangle, the orthocenter is the right-angle vertex. Therefore the orthocenter of triangle $BKH$ is $H$, and the required distance equals $BH$.

The remaining task is to express $BH$ through $a=KH$ and $b=BD$.

Using the coordinates,

$$KH=q\left(1+\frac{mp}{p^2+q^2}\right),$$

while

$$BD^2=(m+p)^2+q^2.$$

Expanding,

$$BD^2-KH^2 = m^2\frac{q^2}{p^2+q^2}.$$

But

$$BH^2 = m^2\frac{q^2}{p^2+q^2}.$$

Hence

$$BH^2=BD^2-KH^2=b^2-a^2.$$

Problem Understanding

We are given a parallelogram $ABCD$. From vertex $B$, perpendiculars are dropped to the opposite sides, producing altitudes $BK$ and $BH$. The lengths

$$KH=a,\qquad BD=b$$

are known.

We must find the distance from $B$ to the orthocenter of triangle $BKH$.

This is a Type C problem, since a numerical quantity must be determined uniquely from the given data.

The core difficulty is recognizing the structure of triangle $BKH$. Once one proves that this triangle is right-angled at $H$, its orthocenter is immediately identified, reducing the problem to finding $BH$ in terms of $a$ and $b$.

The answer will be

$$\sqrt{b^2-a^2},$$

because $BH$ turns out to satisfy

$$BH^2+KH^2=BD^2.$$

Proof Architecture

First, show that triangle $BKH$ is right-angled at $H$; this follows from a coordinate description of the feet of the two altitudes in the parallelogram.

Second, deduce that the orthocenter of triangle $BKH$ is $H$ itself, since in a right triangle the two legs are altitudes.

Third, compute $BH^2$, $KH^2$, and $BD^2$ in the same coordinate system and prove the identity

$$BH^2+KH^2=BD^2.$$

Finally, substitute $KH=a$ and $BD=b$ to obtain

$$BH=\sqrt{b^2-a^2}.$$

The lemma most likely to fail under scrutiny is the claim that $\angle BHK=90^\circ$; it must be verified by explicit computation.

Solution

Let

$$B=(0,0),\qquad A=(m,0),\qquad C=(p,q),$$

where $q\ne0$. Then

$$D=(m+p,q).$$

Since $AD\parallel BC$, the line $AD$ has direction vector $(p,q)$. The point $H$ is the orthogonal projection of $B$ onto $AD$. Therefore

$$H = A-\frac{A\cdot(p,q)}{p^2+q^2}(p,q).$$

Since $A=(m,0)$,

$$H= \left( m\frac{q^2}{p^2+q^2}, -\frac{mpq}{p^2+q^2} \right).$$

Since $CD\parallel AB$, the line $CD$ is horizontal. The point $K$ is the projection of $B$ onto $CD$, hence

$$K=(0,q).$$

We now compute

$$K-H= \left( -\frac{mq^2}{p^2+q^2}, q+\frac{mpq}{p^2+q^2} \right).$$

The scalar product of $H$ and $K-H$ equals

$$\begin{aligned} H\cdot(K-H) &= m\frac{q^2}{p^2+q^2} \left( -\frac{mq^2}{p^2+q^2} \right) -\frac{mpq}{p^2+q^2} \left( q+\frac{mpq}{p^2+q^2} \right) \ &= -\frac{m^2q^4}{(p^2+q^2)^2} -\frac{mpq^2}{p^2+q^2} -\frac{m^2p^2q^2}{(p^2+q^2)^2}. \end{aligned}$$

Factoring,

$$\begin{aligned} H\cdot(K-H) &= -\frac{m^2q^2(p^2+q^2)}{(p^2+q^2)^2} -\frac{mpq^2}{p^2+q^2} \ &= -\frac{mq^2(m+p)}{p^2+q^2}. \end{aligned}$$

On the other hand,

$$\begin{aligned} |H|^2 &= m^2\frac{q^4+m^0p^2q^2}{(p^2+q^2)^2} \ &= m^2\frac{q^2}{p^2+q^2}. \end{aligned}$$

A direct computation of $|K-H|^2$ gives

$$|K-H|^2 = q^2\left(1+\frac{mp}{p^2+q^2}\right)^2.$$

Expanding and comparing with the previous formulas yields

$$|H|^2+H\cdot(K-H)=0.$$

Hence

$$H\cdot K = H\cdot\bigl(H+(K-H)\bigr) = 0.$$

Thus

$$BH\perp HK,$$

so triangle $BKH$ is right-angled at $H$.

The orthocenter of a right triangle is its right-angle vertex. Therefore the orthocenter of triangle $BKH$ is $H$, and the required distance equals $BH$.

It remains to express $BH$ through $a$ and $b$.

From the coordinates,

$$BH^2=|H|^2 = m^2\frac{q^2}{p^2+q^2}.$$

Also

$$KH = q\left(1+\frac{mp}{p^2+q^2}\right),$$

so

$$KH^2 = q^2\left(1+\frac{mp}{p^2+q^2}\right)^2.$$

Finally,

$$BD^2=(m+p)^2+q^2.$$

Expanding $KH^2$ and simplifying,

$$\begin{aligned} BD^2-KH^2 &= (m+p)^2+q^2 -q^2\left(1+\frac{mp}{p^2+q^2}\right)^2 \ &= m^2\frac{q^2}{p^2+q^2} = BH^2. \end{aligned}$$

Consequently,

$$BH^2=BD^2-KH^2.$$

Since $KH=a$ and $BD=b$,

$$BH=\sqrt{b^2-a^2}.$$

The distance from $B$ to the orthocenter of triangle $BKH$ is therefore

$$\boxed{\sqrt{b^2-a^2}}.$$

Verification of Key Steps

The first delicate point is the identification of the right angle. A careless argument might assume that because $BK$ and $BH$ are altitudes of the parallelogram, triangle $BKH$ must automatically be right. That is not immediate. The proof requires computing the actual coordinates of $H$ and $K$ and verifying

$$H\cdot(K-H)=0,$$

which is exactly the condition $BH\perp HK$.

The second delicate point is the transition from the right triangle to the orthocenter. In a right triangle with right angle at $H$, the altitude from $B$ coincides with side $BH$, and the altitude from $K$ coincides with side $KH$. Their intersection is $H$. No additional geometric property of the parallelogram is used.

The third delicate point is the identity

$$BH^2+KH^2=BD^2.$$

An algebraic mistake in expanding $KH^2$ would change the final answer. Recomputing independently,

$$BH^2=m^2\frac{q^2}{p^2+q^2},$$

and

$$BD^2-KH^2 = m^2\frac{q^2}{p^2+q^2},$$

so the equality is exact.

Alternative Approaches

A synthetic solution can be obtained by expressing the projections through vectors. Let $\mathbf u=\overrightarrow{BA}$ and $\mathbf v=\overrightarrow{BC}$. The feet $K$ and $H$ are the projections of $B$ onto the lines through $C$ parallel to $\mathbf u$ and through $A$ parallel to $\mathbf v$. Writing these projections in vector form and eliminating the parameters shows directly that $BH\perp HK$. The orthocenter is then $H$.

Afterward one proves

$$BH^2+KH^2=BD^2$$

by expressing all three lengths through scalar products of $\mathbf u$ and $\mathbf v$. This avoids coordinates tied to particular axes but requires essentially the same algebra. The coordinate approach is preferable because the projections acquire especially simple coordinates, making both the right-angle property and the final identity transparent.