Kvant Math Problem 1271
Let the semicircle have radius $R$ and center $O$.
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Verdicts: SKIP + SKIP
Solve time: 3m49s
Source on kvant.digital
Problem
A semicircle with diameter $AB$ is given. Construct a chord $MN$ parallel to $AB$ such that the trapezoid $AMNB$ is circumscribed.
V. A. Senderov
Exploration
Let the semicircle have radius $R$ and center $O$. Since $MN\parallel AB$, the trapezoid $AMNB$ is isosceles. A trapezoid is circumscribed if and only if the sums of lengths of opposite sides are equal. Thus for our trapezoid the condition becomes
$$AB+MN=AM+BN.$$
Because the trapezoid is isosceles, $AM=BN$, hence
$$AB+MN=2AM.$$
The unknown chord is determined by its distance $h$ from the diameter. Writing everything in coordinates should reveal the required relation.
Take
$$O=(0,0),\qquad A=(-R,0),\qquad B=(R,0).$$
Let the chord be the line $y=h$, where $0<h<R$. Then
$$M=(-\sqrt{R^2-h^2},h),\qquad N=(\sqrt{R^2-h^2},h).$$
The chord length is
$$MN=2\sqrt{R^2-h^2}.$$
Also
$$AM^2=(R-\sqrt{R^2-h^2})^2+h^2 =2R^2-2R\sqrt{R^2-h^2}.$$
Hence
$$AM=\sqrt{2R\bigl(R-\sqrt{R^2-h^2}\bigr)}.$$
Substituting into
$$2R+2\sqrt{R^2-h^2}=2AM$$
and simplifying may be unpleasant. Introduce
$$x=\sqrt{R^2-h^2}.$$
Then
$$AM^2=2R(R-x).$$
The tangential condition gives
$$R+x=AM.$$
Squaring,
$$(R+x)^2=2R(R-x).$$
This yields
$$R^2+2Rx+x^2=2R^2-2Rx,$$
so
$$x^2+4Rx-R^2=0.$$
Since $x>0$,
$$x=R(\sqrt5-2).$$
Thus the chord must be at distance
$$h=\sqrt{R^2-x^2} =R\sqrt{1-(\sqrt5-2)^2}.$$
A direct construction from this formula is unattractive. The important observation is
$$\frac{x}{R}=\sqrt5-2.$$
Since
$$(\sqrt5-2)(\sqrt5+2)=1,$$
we have
$$\frac{R}{x}=\sqrt5+2=\varphi^3,$$
where $\varphi=\frac{1+\sqrt5}{2}$.
So the required half-chord length $x$ satisfies
$$x=\frac{R}{\varphi^3}.$$
The problem becomes a pure ruler and compass construction: on the diameter construct a segment equal to $R/\varphi^3$, then draw through its endpoint the chord parallel to $AB$. The crucial point is proving rigorously that this value indeed comes from the tangential condition.
Problem Understanding
A semicircle with diameter $AB$ is given. We must construct a chord $MN$ parallel to $AB$ such that the trapezoid $AMNB$ admits an inscribed circle.
This is a Type D problem. We must describe an explicit construction and verify that the resulting trapezoid is circumscribed.
The core difficulty is converting the tangency condition for the trapezoid into a geometric condition determining the position of the chord. After that, one must express the resulting distance from the center in a constructible form.
The answer is the unique chord whose half-length equals $R(\sqrt5-2)$, where $R=\frac{AB}{2}$. Intuitively, the tangency condition forces a specific relation between a leg and the two bases, and this relation determines a unique level parallel to the diameter.
Proof Architecture
The first lemma states that a trapezoid $AMNB$ with $MN\parallel AB$ is circumscribed if and only if $AB+MN=2AM$. This follows from the characterization of tangential quadrilaterals and the equality $AM=BN$.
The second lemma states that if the half-length of the chord is $x$, then the tangency condition is equivalent to $(R+x)^2=2R(R-x)$. This comes from coordinate computation of $AM$.
The third lemma states that the unique positive solution is $x=R(\sqrt5-2)$. This follows from solving a quadratic equation.
The final step constructs on the diameter a segment of length $R(\sqrt5-2)=R/\varphi^3$, draws the corresponding parallel chord, and verifies the tangency condition using the preceding lemmas.
The lemma most likely to fail under scrutiny is the derivation of the equation $(R+x)^2=2R(R-x)$, because it encodes the entire geometric condition.
Solution
Let $O$ be the center of the semicircle and let
$$R=OA=OB.$$
Consider an arbitrary chord $MN$ parallel to $AB$. Since the semicircle is symmetric with respect to the perpendicular through $O$, the trapezoid $AMNB$ is isosceles, hence
$$AM=BN.$$
A quadrilateral is circumscribed if and only if the sums of lengths of opposite sides are equal. Applied to $AMNB$, this gives
$$AB+MN=AM+BN.$$
Using $AM=BN$, we obtain
$$AB+MN=2AM.$$
Let
$$x=\frac{MN}{2}.$$
Since the chord is parallel to the diameter, its endpoints have coordinates
$$M=(-x,h),\qquad N=(x,h)$$
for some $h$, and the relation
$$x^2+h^2=R^2$$
holds.
Choose coordinates
$$O=(0,0),\quad A=(-R,0),\quad B=(R,0).$$
Then
$$AM^2=(R-x)^2+h^2.$$
Substituting $h^2=R^2-x^2$ gives
$$AM^2=(R-x)^2+R^2-x^2 =2R(R-x).$$
The condition for a circumscribed trapezoid becomes
$$2R+2x=2AM,$$
or
$$R+x=AM.$$
Squaring and using the expression for $AM^2$,
$$(R+x)^2=2R(R-x).$$
Expanding,
$$R^2+2Rx+x^2=2R^2-2Rx,$$
hence
$$x^2+4Rx-R^2=0.$$
The positive root is
$$x=R(\sqrt5-2).$$
Therefore the required chord is uniquely determined by the condition
$$\frac{MN}{2}=R(\sqrt5-2).$$
Since
$$(\sqrt5-2)(\sqrt5+2)=1,$$
we have
$$x=\frac{R}{\sqrt5+2}.$$
Let
$$\varphi=\frac{1+\sqrt5}{2}.$$
Because
$$\varphi^3=2+\sqrt5,$$
it follows that
$$x=\frac{R}{\varphi^3}.$$
Construction:
Construct the golden ratio segment $\varphi$ from the given segment $R$. Repeatedly construct the segments $R/\varphi$, $R/\varphi^2$, and $R/\varphi^3$. Denote by $x$ the resulting segment $R/\varphi^3$.
On the diameter $AB$, mark points $P$ and $Q$ such that
$$OP=OQ=x,$$
with $P$ and $Q$ on opposite sides of $O$. Through $P$ and $Q$ draw the perpendiculars to $AB$; they meet the semicircle at $M$ and $N$. Since $OP=OQ$, the chord $MN$ is parallel to $AB$ and has half-length
$$\frac{MN}{2}=x=R(\sqrt5-2).$$
By the computation above, this value of $x$ satisfies
$$(R+x)^2=2R(R-x),$$
hence
$$R+x=AM.$$
Since $AB=2R$ and $MN=2x$,
$$AB+MN=2(R+x)=2AM=AM+BN.$$
The criterion for tangential quadrilaterals shows that $AMNB$ is circumscribed.
Thus the constructed chord is the required one.
$$\boxed{\text{The required chord is the unique chord parallel to }AB\text{ whose half-length equals }R(\sqrt5-2)=R/\varphi^3.}$$
Verification of Key Steps
The first delicate step is the use of the tangential quadrilateral criterion. For a circumscribed quadrilateral, opposite sides satisfy
$$AB+MN=AM+BN.$$
Because $MN\parallel AB$, the trapezoid is isosceles, so $AM=BN$. Substituting yields
$$AB+MN=2AM.$$
Without proving $AM=BN$, one would obtain an incomplete condition.
The second delicate step is the computation of $AM$. Starting from
$$AM^2=(R-x)^2+h^2,$$
the circle relation
$$x^2+h^2=R^2$$
gives
$$AM^2=R^2-2Rx+x^2+R^2-x^2 =2R(R-x).$$
A sign error in expanding $(R-x)^2$ would change the quadratic equation and produce an incorrect chord.
The third delicate step is selecting the correct root. The equation
$$x^2+4Rx-R^2=0$$
has roots
$$x=R(-2\pm\sqrt5).$$
Since $x$ is a length, $x>0$, so only
$$x=R(\sqrt5-2)$$
is admissible. The other root is negative and has no geometric meaning.
Alternative Approaches
A synthetic approach avoids coordinates. Let $T$ be the point where the incircle touches the leg $AM$. Tangency gives
$$AT=R,$$
because the tangency lengths from $A$ to the two adjacent sides are equal and the distance from $A$ to the touchpoint on $AB$ equals the inradius contact segment determined by the semicircle. Similar considerations on both legs lead to a relation between the leg length and the upper base. After expressing the geometry through similar right triangles, one arrives at the same quadratic equation for the half-chord length.
The coordinate method is preferable because every quantity involved in the tangency criterion becomes an explicit algebraic expression. The determination of the chord then reduces to a single quadratic equation whose positive solution immediately yields a ruler and compass construction.