Kvant Math Problem 377

The required condition is

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Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 12m03s
Source on kvant.digital

Problem

Given a triangle $ABC$. Find a point $D$ on the side $AC$ such that the perimeter of the triangle $ABD$ equals the length of the side $BC$.

S. Okhitin

Exploration

The required condition is

$AB+BD+AD=BC.$

Rearranging,

$AD+BD=BC-AB.$

A necessary condition for the existence of a solution is

$BC-AB\ge AB,$

because in every triangle $ABD$,

$AD+BD\ge AB.$

Hence any solution must satisfy

$BC\ge 2AB.$

Assume from now on that

$BC\ge 2AB.$

Then $BC-AB\ge0$, so we may choose a point $E$ on the side $AC$ such that

$AE=BC-AB.$

Since

$BC<AB+AC,$

we have

$BC-AB<AC,$

hence $E$ indeed lies on the segment $AC$.

The condition to be achieved becomes

$AD+BD=AE.$

If $D$ lies on the segment $AE$, then

$AE=AD+DE,$

so

\iff AD+BD=AD+DE \iff BD=DE.$$Thus the problem is reduced to finding a point $D$ on the segment $AE$ such that$$BD=DE.$$## Construction Construct the point $E$ on the side $AC$ by$$AE=BC-AB.$$Construct the perpendicular bisector of the segment $BE$. Let $D$ be the intersection of this perpendicular bisector with the segment $AE$. ## Verification Since $D$ lies on the perpendicular bisector of $BE$,$$BD=DE.$$Since $D$ lies on the segment $AE$,$$AE=AD+DE.$$Substituting $DE=BD$ gives$$AE=AD+BD.$$Using the definition of $E$,$$AD+BD=BC-AB.$$Adding $AB$ to both sides yields$$AB+AD+BD=BC.$$Hence the perimeter of triangle $ABD$ is equal to the length of side $BC$. ## Verification of Existence It remains to show that the perpendicular bisector of $BE$ actually meets the segment $AE$ when$$BC\ge2AB.$$Consider the continuous function$$f(X)=XB-XE$$for points $X$ on the segment $AE$. At the endpoint $E$,$$f(E)=EB>0.$$At the endpoint $A$,$$f(A)=AB-AE.$$Because$$AE=BC-AB,$$the condition$$BC\ge2AB$$is equivalent to$$AE\ge AB,$$hence$$f(A)=AB-AE\le0.$$Thus $f(A)\le0<f(E)$. By continuity, there exists a point $D$ on the segment $AE$ for which$$f(D)=0,$$that is,$$DB=DE.$$Every point satisfying $DB=DE$ lies on the perpendicular bisector of $BE$, so this point $D$ is exactly the intersection of that perpendicular bisector with the segment $AE$. Thus the construction is possible whenever$$BC\ge2AB.$$For the converse, if$$BC<2AB,$$then for every point $D$ on $AC$,$$AD+BD\ge AB,$$and therefore$$AB+AD+BD\ge2AB>BC.$$Hence no solution can exist. Consequently, a solution exists if and only if$$BC\ge2AB,$$and when this condition holds, it is obtained by the construction above.