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# Acute angled Triangle | PRMO II 2019 | Question 29

Try this beautiful problem from the Pre-RMO, 2017, Question 23, based on Solving Equation. You may use sequential hints to solve the problem.

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.

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