Try this beautiful Problem on Geometry based on Interior Point of a Triangle from PRMO-2017
Let $P$ be an interior point of a triangle $A B C$ whose side lengths are $26,65,78 .$ The line through P parallel to BC meets AB in K and AC in L. The line through P parallel to CA meets BC in M and BA in N. The line through P parallel to AB meets CA in S and CB in T. If KL, MN, ST are of equal lengths, find this common length.
,
Geometry
Triangle
modulus
Pre College Mathematics
Prmo-2017, Problem-24
no solution is possible.
We have to find out the common length .
Clearly PKBT, PMCL and PSAN are parallelograms. Let $P T=K B=x, P M=L C=y$,
$\mathrm{PK}=\mathrm{BT}=\mathrm{z}$ and $\mathrm{KL}=\mathrm{MN}=\mathrm{ST}=\ell$
because $\Delta P T M \sim \Delta A B C$ [as $PT ||AB$,$PM||AC$, therefore interior angles are same, so $\triangle PTM \sim \triangle ABC$]
Since $\Delta \mathrm{PTM} \sim \Delta \mathrm{ABC} $
$\frac{y}{65}=\frac{26-\ell}{26}$
Again, $\Delta \mathrm{NKP} \sim \Delta \mathrm{ABC}$ [ as $NP||AC$, $KP||BC$, therefore interior angles are same, so $\Delta \mathrm{NKP} \sim \Delta \mathrm{ABC}$ ]
$\Rightarrow \frac{\ell-y}{65}=\frac{78-\ell}{78}$
Adding (1)$\&(2)$
$\frac{\ell}{65}=\frac{26-\ell}{26}+\frac{78-\ell}{78}$
$=\frac{78-3(+78-\ell}{78}$
$\Rightarrow 6 \ell=5(156-4 \ell)$
$\Rightarrow 26 \ell=5 \times 156$
$\Rightarrow \ell=\frac{5 \times 156}{26}=30$
But $\ell$ must be less than $26,$ hence no solution is possible.
Try this beautiful Problem on Geometry based on Interior Point of a Triangle from PRMO-2017
Let $P$ be an interior point of a triangle $A B C$ whose side lengths are $26,65,78 .$ The line through P parallel to BC meets AB in K and AC in L. The line through P parallel to CA meets BC in M and BA in N. The line through P parallel to AB meets CA in S and CB in T. If KL, MN, ST are of equal lengths, find this common length.
,
Geometry
Triangle
modulus
Pre College Mathematics
Prmo-2017, Problem-24
no solution is possible.
We have to find out the common length .
Clearly PKBT, PMCL and PSAN are parallelograms. Let $P T=K B=x, P M=L C=y$,
$\mathrm{PK}=\mathrm{BT}=\mathrm{z}$ and $\mathrm{KL}=\mathrm{MN}=\mathrm{ST}=\ell$
because $\Delta P T M \sim \Delta A B C$ [as $PT ||AB$,$PM||AC$, therefore interior angles are same, so $\triangle PTM \sim \triangle ABC$]
Since $\Delta \mathrm{PTM} \sim \Delta \mathrm{ABC} $
$\frac{y}{65}=\frac{26-\ell}{26}$
Again, $\Delta \mathrm{NKP} \sim \Delta \mathrm{ABC}$ [ as $NP||AC$, $KP||BC$, therefore interior angles are same, so $\Delta \mathrm{NKP} \sim \Delta \mathrm{ABC}$ ]
$\Rightarrow \frac{\ell-y}{65}=\frac{78-\ell}{78}$
Adding (1)$\&(2)$
$\frac{\ell}{65}=\frac{26-\ell}{26}+\frac{78-\ell}{78}$
$=\frac{78-3(+78-\ell}{78}$
$\Rightarrow 6 \ell=5(156-4 \ell)$
$\Rightarrow 26 \ell=5 \times 156$
$\Rightarrow \ell=\frac{5 \times 156}{26}=30$
But $\ell$ must be less than $26,$ hence no solution is possible.
