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Try this beautiful problem from the Pre-RMO II, 2019, Question 29, based on Acute angled triangle.

## Acute angled triangle – Problem 29

Let ABC be a acute angled triangle with AB=15 and BC=8. Let D be a point on AB such that BD=BC. Consider points E on AC such that $\angle$DEB=$\angle$BEC. If $\alpha$ denotes the product of all possible val;ues of AE, find[$\alpha$] the integer part of $\alpha$.

• is 107
• is 68
• is 840
• cannot be determined from the given information

### Key Concepts

Equation

Algebra

Integers

But try the problem first…

Source

PRMO II, 2019, Question 29

Higher Algebra by Hall and Knight

## Try with Hints

First hint

The pairs $E_1$,$E_2$ satisfies condition or $E_1$=intersection of CBO with AC and $E_2$=intersection of $\angle$bisector of B and AC

since that $\angle DE_2B$=$\angle CE_2B$ and for $E_1$$\angle BE_1C$=$\angle$BDC=$\angle$BCD=$\angle BE_1D$

or, $AE_1.AC$=$AD.AB$=$7 \times 15$

$\frac{AE_2}{AC}$=$\frac{XY}{XC}$

(for y is midpoint of OC and X is foot of altitude from A to CD)

Second Hint

$\frac{XD}{DY}=\frac{7}{8}$ and DY=YC

or, $\frac{XD+DY}{XC}$=$\frac{15}{7+8+8}$=$\frac{15}{23}$

or, $\frac{XY}{XC}=\frac{15}{23}$

or, $\frac{AE_2}{AC}$=$\frac{15}{23}$

or, $AE_1.AE_2$=$\frac{15}{23}(7.15)$=$\frac{225 \times 7}{23}$

Final Step

$[\frac{225 \times 7}{23}]$=68.