Daily · 2026-05-08

Three problems · Geometry, Compound Interest, Algebra

One bite-sized math problem set for the day. Read the statement, think it through, then expand the solution to check your reasoning.

#1 · Geometry·Medium

A right triangle has legs of length 3 and 4. What is the radius of its inscribed circle?

A. $1$
B. $\dfrac{6}{5}$
C. $\dfrac{5}{2}$
D. $\sqrt{2}$

Solution

For any triangle, the inradius is

r = A / s

, where

A

is the area and

s

is the semi-perimeter.

Hypotenuse:

\sqrt{3^2 + 4^2} = 5

. Area:

\dfrac{1}{2} \cdot 3 \cdot 4 = 6

. Semi-perimeter:

(3 + 4 + 5)/2 = 6

r = 6 / 6 = 1

.

For right triangles specifically there's also a shortcut:

r = (a + b - c)/2 = (3 + 4 - 5)/2 = 1

Two formulas, same answer. The general

r = A/s

works for any triangle; the right-triangle shortcut

r = (a+b-c)/2

is faster when applicable. The 3-4-5 triangle is special enough that you should expect a clean integer.

#2 · Compound Interest·Medium

You owe \

10,000 on a loan at 6\% annual interest, compounded annually. You pay \

2,000 at the end of each year. After how many years is the loan fully paid off (rounded up)?

A.5 years
B.6 years
C.7 years
D.10 years

Solution

Let

B_n

be the balance after the

n

-th payment.

B_0 = 10000

, and

B_n = 1.06 \cdot B_{n-1} - 2000

B_1 = 10600 - 2000 = 8600

B_2 = 9116 - 2000 = 7116

B_3 = 7542.96 - 2000 = 5542.96

B_4 = 5875.54 - 2000 = 3875.54

B_5 = 4108.07 - 2000 = 2108.07

B_6 = 2234.55 - 2000 = 234.55

After year 6 there's only \

234.55 left, so a 6th payment of \

234.55 finishes it. Total: 6 years (the last payment is small).

The recursion

B_n = (1+r) B_{n-1} - P

is how every amortization schedule works under the hood. There's a closed-form using geometric-sum arithmetic, but for one-off questions stepping through year by year is faster than deriving the formula.

#3 · Algebra·Easy

Simplify:

\quad \dfrac{x^2 - 9}{x^2 - 6x + 9}

for

x \ne 3

A. $\dfrac{x + 3}{x - 3}$
B. $\dfrac{x - 3}{x + 3}$
C. $x - 3$
D. $\dfrac{1}{x - 3}$

Solution

Factor numerator and denominator:

\frac{x^2 - 9}{x^2 - 6x + 9} = \frac{(x-3)(x+3)}{(x-3)^2}

Cancel one factor of

(x - 3)

(valid since

x \ne 3

= \frac{x + 3}{x - 3}

Difference-of-squares on top, perfect-square trinomial on bottom. Two pattern recognitions, then one cancellation. The exclusion

x \ne 3

matters because the original expression is undefined there even though the simplified one *almost* is.

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